Deck 7: Estimates and Sample Size

ملء الشاشة (f)
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سؤال
Bert constructed a confidence interval to estimate the mean weight of students in his class.
The population was very small - only 30. Ruth constructed a confidence interval for the mean weight of all adult males in the city. She based her confidence interval on a very small sample of only 5. Which confidence interval is likely to give a better estimate of the mean it is estimating? Which is likely to be more of a problem, a small sample or a small population?
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سؤال
When determining the sample size for a desired margin of error, the formula is n=[zα/2]2p^q^E2\mathrm { n } = \frac { [ \mathrm { z } \alpha / 2 ] ^ { 2 } \hat { \mathrm { p }}\hat { \mathrm { q } } } { \mathrm { E } ^ { 2 } } . Based on this formula, discuss the fact that sample size is not dependent on the population size; that is, it is not necessary to sample a particular percent of the population.
سؤال
Describe the process for finding the confidence interval for a population proportion.
سؤال
Why would manufacturers and businesses be interested in constructing a confidence interval for the population variance? Would manufacturers and businesses want large or small variances?
سؤال
When determining the sample size needed to achieve a particular error estimate you need to know ?. What are two methods of estimating ? if ? is unknown?
سؤال
 Why is s2 the best point estimate of σ2 ? \text { Why is } \mathrm { s } ^ { 2 } \text { the best point estimate of } \sigma ^ { 2 } \text { ? }
سؤال
Hannah selected a simple random sample of all adults in her town and, based on this sample, constructed a confidence interval for the mean salary of all adults in the town.
However, the distribution of salaries in the town is not exactly normal. Will the confidence interval still give a good estimate of the mean salary?
سؤال
How do you determine whether to use the z or t distribution in computing the margin of  error, E=zα/2σn or E=tα/2sn ? \text { error, } \mathrm { E } = \mathrm { z } _ { \alpha / 2 } \cdot \frac { \sigma } { \sqrt { \mathrm { n } } } \text { or } \mathrm { E } = \mathrm { t } _ { \alpha / 2 } \cdot \frac { \mathrm { s } } { \sqrt { \mathrm { n } } } \text { ? }
سؤال
Describe the steps for finding a confidence interval.
سؤال
Complete the table to compare z and t distributions. Complete the table to compare z and t distributions.  <div style=padding-top: 35px>
سؤال
Interpret the following 95% confidence interval for mean weekly salaries of shift managers at Guiseppe's Pizza and Pasta. 325.80 < µ < 472.30
سؤال
Under what circumstances can you replace ? with s in the formula E=zα/2σn\mathrm { E } = \mathrm { z } _ { \alpha / 2 } \cdot \frac { \sigma } { \sqrt { \mathrm { n } } }
سؤال
A researcher is interested in estimating the proportion of voters who favor a tax on e-commerce. Based on a sample of 250 people, she obtains the following 99% confidence interval for the population proportion, p:0.113<p<0.171\mathrm { p } : 0.113 < \mathrm { p } < 0.171 Which of the statements below is a valid interpretation
Of this confidence interval?

A) If 100 different samples of size 250 were selected and, based on each sample, a confidence interval were constructed, exactly 99 of these confidence intervals would contain the true value of p.
B) If many different samples of size 250 were selected and, based on each sample, a confidence interval were constructed, 99% of the time the true value of p would lie between 0.113 and 0.171.
C) There is a 99% chance that the true value of p lies between 0.113 and 0.171.
D) If many different samples of size 250 were selected and, based on each sample, a confidence interval were constructed, in the long run 99% of the confidence intervals would contain the true value of p.
سؤال
What assumption about the parent population is needed to use the t distribution to compute the margin of error?
سؤال
Based on a simple random sample of students from her school, Sally obtained a point estimate of the mean weight of students at her school. What additional information would be provided by a confidence interval estimate of the mean weight?
سؤال
What is the best point estimate for the population proportion? Explain why that point estimate is best.
سؤال
Explain how confidence intervals might be used to make decisions. Give an example to clarify your explanation.
سؤال
When determining sample size we need to know p^\hat { \mathrm { p } } . If we have no prior information, what are two methods that can be used?
سؤال
 <div style=padding-top: 35px>
سؤال
Explain the difference between descriptive and inferential statistics.
سؤال
50 people are selected randomly from a certain population and it is found that 18 people in the sample are over 6 feet tall. What is the point estimate of the proportion of people in the population who are over 6 feet tall?

A) 0.50
B) 0.36
C) 0.64
D) 0.23
سؤال
A 99%99 \% confidence interval (in inches) for the mean height of a population is 66.2<μ<67.666.2 < \mu < 67.6 . This result is based on a sample of size 144 . Construct the 95%95 \% confidence interval. (Hint: you will first need to find the sample mean and sample standard deviation).

A) 66.266.2 in <μ<67.6< \mu < 67.6 in.
B) 66.066.0 in <μ<67.8< \mu < 67.8 in.
C) 66.366.3 in <μ<67.5< \mu < 67.5 in.
D) 66.466.4 in <μ<67.4in< \mu < 67.4 \mathrm { in } .
سؤال
Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.

-A sociologist develops a test to measure attitudes towards public transportation, and 27 randomly selected subjects are given the test. Their mean score is 76.2 and their standard deviation is 21.4.
Construct the 95% confidence interval for the mean score of all such subjects.

A) 74.6<μ<77.874.6 < \mu < 77.8
B) 67.7<μ<84.767.7 < \mu < 84.7
C) 64.2<μ<88.264.2 < \mu < 88.2
D) 69.2<μ<83.269.2 < \mu < 83.2
سؤال
Use the confidence level and sample data to find a confidence interval for estimating the population µ. Round your answer to the same number of decimal places as the sample mean.

-Test scores: n=79,x=65.0,σ=5.0;98%\mathrm { n } = 79 , \overline { \mathrm { x } } = 65.0 , \sigma = 5.0 ; 98 \% confidence

A) 64.1<μ<65.964.1 < \mu < 65.9
B) 63.9<μ<66.163.9 < \mu < 66.1
C) 63.5<μ<66.563.5 < \mu < 66.5
D) 63.7<μ<66.363.7 < \mu < 66.3
سؤال
The Bide-a-While efficiency hotel, which caters to business workers who stay for extended periods of time (weeks or months), offers room service. In a small study of 35 randomly selected room service orders, the 95% confidence interval for mean delivery time for room service is 24.8<μ<29.624.8 < \mu < 29.6 minutes. The marketing director is trying to determine if she can advertise "room service in under 30 minutes, or the order is free." How would you advise her?
سؤال
Use the confidence level and sample data to find a confidence interval for estimating the population µ. Round your answer to the same number of decimal places as the sample mean.

-A group of 64 randomly selected students have a mean score of 38.638.6 with a standard deviation of 4.94.9 on a placement test. What is the 90%90 \% confidence interval for the mean score, μ\mu , of all students taking the test?

A) 37.4<μ<39.837.4 < \mu < 39.8
B) 37.2<μ<40.037.2 < \mu < 40.0
C) 37.6<μ<39.637.6 < \mu < 39.6
D) 37.0<μ<40.237.0 < \mu < 40.2
سؤال
Under what three conditions is it appropriate to use the t distribution in place of the standard normal distribution?
سؤال
Use the given data to find the minimum sample size required to estimate the population proportion.

-Margin of error: 0.018; confidence level: 99%;p^ and q^ unknown 99 \% ; \hat { \mathrm { p } } \text { and } \hat { \mathrm { q } } \text { unknown }

A) 4966
B) 4114
C) 7116
D) 5117
سؤال
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.

- n=110,x=68;88%\mathrm { n } = 110 , \mathrm { x } = 68 ; 88 \% confidence

A) 0.545<p<0.6910.545 < p < 0.691
B) 0.541<p<0.6950.541 < p < 0.695
C) 0.542<p<0.6940.542 < p < 0.694
D) 0.546<p<0.6900.546 < p < 0.690
سؤال
In constructing a confidence interval for σ\sigma or σ2\sigma ^ { 2 } , a table is used to find the critical values χL2\chi _ { L } ^ { 2 } and χ2R\chi \underset { R } { 2 } for values of n101n \leq 101 . For larger values of n,χL2n , \chi _ { L } ^ { 2 } and χR2\chi { } _ { \mathrm { R } } ^ { 2 } can be approximated by using the following formula: χ2=12[±zα/2+2k1]2\chi ^ { 2 } = \frac { 1 } { 2 } [ \pm \mathrm { z } \alpha / 2 + \sqrt { 2 \mathrm { k } - 1 } ] ^ { 2 } where k\mathrm { k } is the number of degrees of freedom and zα/2\mathrm { z } \alpha / 2 is the critical z score. Construct the 90%90 \% confidence interval for σ\sigma using the following sample data: a sample of size n=234\mathrm { n } = 234 yields a mean weight of 155lb155 \mathrm { lb } and a standard deviation of 26.0lb26.0 \mathrm { lb } . Round the confidence interval limits to the nearest hundredth.

A) 24.13lb<σ<28.11lb24.13 \mathrm { lb } < \sigma < 28.11 \mathrm { lb }
B) 24.18lb<σ<28.18lb24.18 \mathrm { lb } < \sigma < 28.18 \mathrm { lb }
C) 24.57lb<σ<27.67lb24.57 \mathrm { lb } < \sigma < 27.67 \mathrm { lb }
D) 23.86lb<σ<28.63lb23.86 \mathrm { lb } < \sigma < 28.63 \mathrm { lb }
سؤال
A radio show host asked people to call in and say whether they support new legislation to promote cleaner sources of energy. Based on this sample, she constructed a confidence interval to estimate the proportion of all listeners to her show who support the legislation.
Is the confidence interval likely to give a good estimate of the proportion of her listeners who support the legislation?
سؤال
A paper published the results of a poll. It stated that, based on a sample of 1000 married men, 51% of married men say that they would marry the same woman again. The margin of error was given as ±3 percentage points and the confidence level was given as 95%.
What does it mean that the margin of error was ±3 percentage points?
سؤال
Use the given data to find the minimum sample size required to estimate the population proportion.

-Margin of error: 0.008; confidence level: 99%; from a prior study, p^\hat{p} is estimated by 0.208.

A) 9889
B) 15,361
C) 137
D) 17,068
سؤال
Define a point estimate. What is the best point estimate for Define a point estimate. What is the best point estimate for   ?<div style=padding-top: 35px> ?
سؤال
Mark wanted to estimate the mean number of years of education of adults in his city. He waited outside a public library and interviewed every tenth adult leaving. Based on this sample, he constructed a confidence interval for the mean number of years of education of adults in the city. Do you think this confidence interval will give a good estimate? Why or why not?
سؤال
Draw a diagram of the chi-square distribution. Discuss its shape and values.
سؤال
Define confidence interval and degree of confidence. Make up an example of a confidence interval and interpret the result.
سؤال
Of 366 randomly selected medical students, 27 said that they planned to work in a rural community. Find a 95% confidence interval for the true proportion of all medical students who plan to work in a rural community.

A) 0.0513<p<0.09620.0513 < p < 0.0962
B) 0.0470<p<0.1010.0470 < p < 0.101
C) 0.0419<p<0.1060.0419 < p < 0.106
D) 0.0386<p<0.1090.0386 < p < 0.109
سؤال
Use the given data to find the minimum sample size required to estimate the population proportion.

-Margin of error: 0.0280.028 ; confidence level: 99%;p^99 \% ; \hat { \mathrm { p } } and q^\hat { \mathrm { q } } unknown

A) 2223
B) 1116
C) 2115
D) 1939
سؤال
Identify the correct distribution (z, t, or neither) for each of the following.  Sample Size  Standard  deviation  Shape of the  distribution  z or t or neither n=35s=4.5 Somewhat skewed n=20s=4.5 Bell shaped n=25σ=4.5 Bell shaped n=20σ=4.5 Extremely skewed \begin{array} { | l | l | l | l | } \hline \text { Sample Size } & \begin{array} { l } \text { Standard } \\\text { deviation }\end{array} & \begin{array} { l } \text { Shape of the } \\\text { distribution }\end{array} & \text { z or t or neither } \\\hline n = 35 & s = 4.5 & \text { Somewhat skewed } & \\\hline n = 20 & s = 4.5 & \text { Bell shaped } & \\\hline n = 25 & \sigma = 4.5 & \text { Bell shaped } & \\\hline n = 20 & \sigma = 4.5 & \text { Extremely skewed } & \\\hline\end{array}
سؤال
Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.

- 90% confidence; n=430,x=8090 \% \text { confidence; } n = 430 , x = 80

A) 0.0386
B) 0.0368
C) 0.0331
D) 0.0309
سؤال
A newspaper article about the results of a poll states: "In theory, the results of such a poll, in 99 cases out of 100 should differ by no more than 2 percentage points in either direction from what would have been obtained by interviewing all voters in the United States." Find the sample size suggested by this statement.

A) 2402
B) 4145
C) 165
D) 3394
سؤال
Do one of the following, as appropriate: (a) Find the critical value z?/2, (b) find the critical value t?/2, (c) state that neither the normal nor the t distribution applies.

- 90%;n=10;σ90 \% ; \mathrm { n } = 10 ; \sigma is unknown; population appears to be normally distributed.

A) tα/2=1.833\mathrm { t } _ { \alpha / 2 } = 1.833
B) zα/2=2.262\mathrm { z } _ { \alpha / 2 } = 2.262
C) tα/2=1.812\mathrm { t } _ { \alpha / 2 } = 1.812
D) zα/2=1.383\mathrm { z } _ { \alpha / 2 } = 1.383
سؤال
In a clinical test with 8900 subjects, 4450 showed improvement from the treatment. Find the margin of error for the 99% confidence interval used to estimate the population proportion.

A) 0.00780
B) 0.0137
C) 0.0120
D) 0.0104
سؤال
Use the given information to find the minimum sample size required to estimate an unknown population mean µ.

-  Margin of error: $134, confidence level: 95%,σ=$575\text { Margin of error: } \$ 134 \text {, confidence level: } 95 \% , \sigma = \$ 575

A) 71
B) 62
C) 50
D) 100
سؤال
Use the given data to find the minimum sample size required to estimate the population proportion.

-  Margin of error: 0.008; confidence level: 99%;p^ and q^ unknown \text { Margin of error: } 0.008 \text {; confidence level: } 99 \% ; \hat { \mathrm { p } } \text { and } \hat { \mathrm { q } } \text { unknown }

A) 25,901
B) 15,900
C) 26,024
D) 25,894
سؤال
Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation ?.
Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation.

-Weights of men: 90%90 \% confidence; n=14,x=154.8lb,s=11.1lb\mathrm { n } = 14 , \overline { \mathrm { x } } = 154.8 \mathrm { lb } , \mathrm { s } = 11.1 \mathrm { lb }

A) 8.7lb<σ<14.3lb8.7 \mathrm { lb } < \sigma < 14.3 \mathrm { lb }
B) 8.2lb<σ<15.6lb8.2 \mathrm { lb } < \sigma < 15.6 \mathrm { lb }
C) 9.0lb<σ<2.7lb9.0 \mathrm { lb } < \sigma < 2.7 \mathrm { lb }
D) 8.5lb<σ<16.5lb8.5 \mathrm { lb } < \sigma < 16.5 \mathrm { lb }
سؤال
Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.

-The principal randomly selected six students to take an aptitude test. Their scores were: 81.478.671.377.876.977.7\begin{array} { l l l l l l } 81.4 & 78.6 & 71.3 & 77.8 & 76.9 & 77.7 \end{array}
Determine a 90%90 \% confidence interval for the mean score for all students.

A) 80.11<μ<74.4580.11 < \mu < 74.45
B) 80.01<μ<74.5580.01 < \mu < 74.55
C) 74.55<μ<80.0174.55 < \mu < 80.01
D) 74.45<μ<80.1174.45 < \mu < 80.11
سؤال
Solve the problem. Round the point estimate to the nearest thousandth.
364 randomly selected light bulbs were tested in a laboratory, 124 lasted more than 500 hours. Find a point estimate of the proportion of all light bulbs that last more than 500 hours.

A) 0.254
B) 0.659
C) 0.338
D) 0.341
سؤال
Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation ?.
Assume that the population has a normal distribution. Round the confidence interval limits to one more decimal place than is used for the original set of data.

-The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were:
7913675951513\begin{array} { c c c c c } 7 & 9 & 13 & 6 & 7 \\5 & 9 & 5 & 15 & 13\end{array}
Find a 95%95 \% confidence interval for the population standard deviation σ\sigma .

A) 0.8 min<σ<2.4 min0.8 \mathrm {~min} < \sigma < 2.4 \mathrm {~min}
B) 2.4 min<σ<6.0 min2.4 \mathrm {~min} < \sigma < 6.0 \mathrm {~min}
C) 2.5 min<σ<6.0 min2.5 \mathrm {~min} < \sigma < 6.0 \mathrm {~min}
D) 2.5 min<σ<6.6 min2.5 \mathrm {~min} < \sigma < 6.6 \mathrm {~min}
سؤال
Use the given data to find the minimum sample size required to estimate the population proportion.

-Margin of error: 0.090.09 ; confidence level: 95%95 \% ; from a prior study, p^\hat { \mathrm { p } } is estimated by the decima equivalent of 87%87 \% .

A) 5
B) 48
C) 54
D) 162
سؤال
Find the critical value χR2\chi { } _ { \mathrm { R } } ^ { 2 } corresponding to a sample size of 19 and a confidence level of 99 percent.

A) 34.80534.805
B) 37.15637.156
C) 6.2656.265
D) 7.0157.015
سؤال
Use the given information to find the minimum sample size required to estimate an unknown population mean µ.

-How many students must be randomly selected to estimate the mean weekly earnings of students at one college? We want 95% confidence that the sample mean is within $2 of the population mean, and the population standard deviation is known to be $10.

A) 136
B) 97
C) 68
D) 85
سؤال
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.

- n=150,x=54;90%\mathrm { n } = 150 , \mathrm { x } = 54 ; 90 \% confidence

A) 0.299<p<0.4210.299 < p < 0.421
B) 0.296<p<0.4240.296 < p < 0.424
C) 0.298<p<0.4220.298 < p < 0.422
D) 0.294<p<0.4260.294 < p < 0.426
سؤال
Use the given information to find the minimum sample size required to estimate an unknown population mean µ.

-  Margin of error: $136, confidence level: 99%,σ=$584\text { Margin of error: } \$ 136 \text {, confidence level: } 99 \% , \sigma = \$ 584

A) 62
B) 123
C) 50
D) 71
سؤال
Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.

-A savings and loan association needs information concerning the checking account balances of its local customers. A random sample of 14 accounts was checked and yielded a mean balance of $664.14 and a standard deviation of $297.29. Find a 98% confidence interval for the true mean checking account balance for local customers.

A) $492.52<μ<$835.76\$ 492.52 < \mu < \$ 835.76
B) $493.71<μ<$834.57\$ 493.71 < \mu < \$ 834.57
C) $453.59<μ<$874.69\$ 453.59 < \mu < \$ 874.69
D) $455.65<μ<$872.63\$ 455.65 < \mu < \$ 872.63
سؤال
Use the given data to find the minimum sample size required to estimate the population proportion.

-  Margin of error: 0.009; confidence level: 99%;p^ and q^ unknown \text { Margin of error: 0.009; confidence level: } 99 \% ; \hat { \mathrm { p } } \text { and } \hat { \mathrm { q } } \text { unknown }

A) 20,465
B) 21,442
C) 9642
D) 19,566
سؤال
Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.

-90% confidence; the sample size is 1410, of which 40% are successes

A) 0.0167
B) 0.0267
C) 0.0256
D) 0.0215
سؤال
Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.

-95% confidence; the sample size is 6100, of which 40% are successes

A) 0.00923
B) 0.0141
C) 0.0123
D) 0.0162
سؤال
Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.

- 95% confidence; n=470,x=5095 \% \text { confidence; } \mathrm { n } = 470 , \mathrm { x } = 50

A) 0.0251
B) 0.0335
C) 0.0293
D) 0.0279
سؤال
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.

-Of 80 adults selected randomly from one town, 67 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance.

A) 0.741<p<0.9340.741 < \mathrm { p } < 0.934
B) 0.731<p<0.9440.731 < \mathrm { p } < 0.944
C) 0.757<p<0.9180.757 < p < 0.918
D) 0.770<p<0.9050.770 < p < 0.905
سؤال
Use the confidence level and sample data to find a confidence interval for estimating the population µ. Round your answer to the same number of decimal places as the sample mean.

-A random sample of 108 light bulbs had a mean life of xˉ=547\bar { x } = 547 hours with a standard deviation of σ=36\sigma = 36 hours. Construct a 90%90 \% confidence interval for the mean life, μ\mu , of all light bulbs of this type.

A) 541hr<μ<553hr541 \mathrm { hr } < \mu < 553 \mathrm { hr }
B) 538hr<μ<556hr538 \mathrm { hr } < \mu < 556 \mathrm { hr }
C) 539hr<μ<555hr539 \mathrm { hr } < \mu < 555 \mathrm { hr }
D) 540hr<μ<554hr540 \mathrm { hr } < \mu < 554 \mathrm { hr }
سؤال
Use the given information to find the minimum sample size required to estimate an unknown population mean µ.

-How many women must be randomly selected to estimate the mean weight of women in one age group. We want 90% confidence that the sample mean is within 2.8 lb of the population mean, and the population standard deviation is known to be 27 lb.

A) 358
B) 253
C) 252
D) 250
سؤال
Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.

-Thirty randomly selected students took the calculus final. If the sample mean was 75 and the standard deviation was 13.2, construct a 99% confidence interval for the mean score of all students.

A) 68.38<μ<81.6268.38 < \mu < 81.62
B) 70.91<μ<79.0970.91 < \mu < 79.09
C) 68.36<μ<81.6468.36 < \mu < 81.64
D) 69.07<μ<80.9369.07 < \mu < 80.93
سؤال
The amounts (in ounces) of juice in eight randomly selected juice bottles are:
15.315.315.715.715.315.915.315.9\begin{array} { l l l l l } 15.3 & 15.3 & 15.7 & 15.7\\15.3&15 .9&15 .3&15 .9\\ \end{array}
Construct a 98%98 \% confidence interval for the mean amount of juice in all such bottle

A) 15.83oz<μ<15.27oz15.83 \mathrm { oz } < \mu < 15.27 \mathrm { oz }
B) 15.93oz<μ<15.17oz15.93 \mathrm { oz } < \mu < 15.17 \mathrm { oz }
C) 15.27oz<μ<15.83oz15.27 \mathrm { oz } < \mu < 15.83 \mathrm { oz }
D) 15.17oz<μ<15.93oz15.17 \mathrm { oz } < \mu < 15.93 \mathrm { oz }
سؤال
Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation ?.
Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation.

-A sociologist develops a test to measure attitudes about public transportation, and 27 randomly selected subjects are given the test. Their mean score is 76.2 and their standard deviation is 21.4.
Construct the 95% confidence interval for the standard deviation, ?, of the scores of all subjects.

A) 16.9<σ<29.316.9 < \sigma < 29.3
B) 17.5<σ<27.817.5 < \sigma < 27.8
C) 16.6<σ<28.616.6 < \sigma < 28.6
D) 17.2<σ<27.217.2 < \sigma < 27.2
سؤال
Use the given data to find the minimum sample size required to estimate the population proportion.

-Margin of error: 0.04; confidence level: 99%; from a prior study, p^\hat { p } is estimated by 0.14.

A) 499
B) 20
C) 289
D) 599
سؤال
Express the confidence interval using the indicated format.

-Express the confidence interval 0.307±0.0580.307 \pm 0.058 in the form of p^E<p<p^+E\hat { p } - \mathrm { E } < \mathrm { p } < \hat { \mathrm { p } } + \mathrm { E } .

A) 0.278<p<0.3360.278 < p < 0.336
B) 0.307<p<0.3650.307 < p < 0.365
C) 0.249<p<0.3070.249 < p < 0.307
D) 0.249<p<0.3650.249 < p < 0.365
سؤال
Find the critical value zα/2\mathrm { z } _ { \alpha / 2 } that corresponds to a 98% confidence level.

A) 2.575
B) 2.33
C) 2.05
D) 1.75
سؤال
Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.

-99% confidence; the sample size is 1180, of which 45% are successes

A) 0.0297
B) 0.0284
C) 0.0337
D) 0.0373
سؤال
Find the critical value χR2\chi _ { \mathrm { R } } ^ { 2 } corresponding to a sample size of 7 and a confidence level of 90 percent.

A) 1.635
B) 12.592
C) 18.548
D) 16.812
سؤال
Express the confidence interval using the indicated format.

-Express the confidence interval 0.62<p<0.720.62 < p < 0.72 in the form of p^±E\hat { p } \pm E .

A) 0.67±0.050.67 \pm 0.05
B) 0.62±0.10.62 \pm 0.1
C) 0.62±0.050.62 \pm 0.05
D) 0.67±0.10.67 \pm 0.1
سؤال
 Find zα/2 for α=0.06\text { Find } z _ { \alpha / 2 } \text { for } \alpha = 0.06

A) 1.88
B) 1.555
C) 1.96
D) 2.75
سؤال
Use the confidence level and sample data to find a confidence interval for estimating the population µ. Round your answer to the same number of decimal places as the sample mean.

-Test scores: n=102,x=77.4,σ=6.5;99%\mathrm { n } = 102 , \overline { \mathrm { x } } = 77.4 , \sigma = 6.5 ; 99 \% confidence

A) 76.1<μ<78.776.1 < \mu < 78.7
B) 75.7<μ<79.175.7 < \mu < 79.1
C) 75.9<μ<78.975.9 < \mu < 78.9
D) 76.3<μ<78.576.3 < \mu < 78.5
سؤال
The following confidence interval is obtained for a population proportion, p:0.686<p<0.712p : 0.686 < p < 0.712 . Use these confidence interval limits to find the point estimate, p^\hat {\mathrm { p }} .

A) 0.699
B) 0.704
C) 0.686
D) 0.694
سؤال
Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.

- n=10,x=12.7, s=3.7,95%\mathrm { n } = 10 , \overline { \mathrm { x } } = 12.7 , \mathrm {~s} = 3.7,95 \% confidence

A) 10.07<μ<15.3310.07 < \mu < 15.33
B) 10.05<μ<15.3510.05 < \mu < 15.35
C) 10.09<μ<15.3110.09 < \mu < 15.31
D) 10.56<μ<14.8410.56 < \mu < 14.84
سؤال
n=12,x=21.9, s=4.0,99%\mathrm { n } = 12 , \overline { \mathrm { x } } = 21.9 , \mathrm {~s} = 4.0,99 \% confidence

A) 18.24<μ<25.5618.24 < \mu < 25.56
B) 18.76<μ<25.0418.76 < \mu < 25.04
C) 18.31<μ<25.4918.31 < \mu < 25.49
D) 18.33<μ<25.4718.33 < \mu < 25.47
سؤال
Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.

-A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 198 milligrams with s=10.5\mathrm { s } = 10.5 milligrams. Construct a 95%95 \% confidence interval for the true mean cholesterol content of all such eggs.

A) 191.3mg<μ<204.7mg191.3 \mathrm { mg } < \mu < 204.7 \mathrm { mg }
B) 191.4mg<μ<204.6mg191.4 \mathrm { mg } < \mu < 204.6 \mathrm { mg }
C) 191.2mg<μ<204.8mg191.2 \mathrm { mg } < \mu < 204.8 \mathrm { mg }
D) 192.6mg<μ<203.4mg192.6 \mathrm { mg } < \mu < 203.4 \mathrm { mg }
سؤال
Use the given data to find the minimum sample size required to estimate the population proportion.

-Margin of error: 0.008; confidence level 95%;p^ and q^ unknown 95 \% ; \hat { \mathrm { p } } \text { and } \hat { \mathrm { q } } \text { unknown }

A) 15,007
B) 15,098
C) 14,488
D) 5045
سؤال
Find the chi-square value xL2x _ { L } ^ { 2 } corresponding to a sample size of 4 and a confidence level of 98 percent.

A) 0.216
B) 9.348
C) 0.115
D) 11.345
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Deck 7: Estimates and Sample Size
1
Bert constructed a confidence interval to estimate the mean weight of students in his class.
The population was very small - only 30. Ruth constructed a confidence interval for the mean weight of all adult males in the city. She based her confidence interval on a very small sample of only 5. Which confidence interval is likely to give a better estimate of the mean it is estimating? Which is likely to be more of a problem, a small sample or a small population?
Bert's confidence interval is likely to give a better estimate. A small sample is more likely to be a problem than a small population.
2
When determining the sample size for a desired margin of error, the formula is n=[zα/2]2p^q^E2\mathrm { n } = \frac { [ \mathrm { z } \alpha / 2 ] ^ { 2 } \hat { \mathrm { p }}\hat { \mathrm { q } } } { \mathrm { E } ^ { 2 } } . Based on this formula, discuss the fact that sample size is not dependent on the population size; that is, it is not necessary to sample a particular percent of the population.
As shown in the formula, the appropriate sample size is dependent on the appropriate z score, the sample proportion, and the margin of error, not on N, the population size.
3
Describe the process for finding the confidence interval for a population proportion.
1) Find the summary statistics p^\hat { \mathrm { p } } and q^\hat { \mathrm { q } }
2) Compute E using the z distribution and the formula
E=zα/2pq^n.E = { } _ { z } \alpha / 2 \cdot \sqrt { \frac { \hat { p q } } { \mathrm { n } } } .
3) Find the interval by adding EE to p^\hat { p } and then subtracting EE from p^\hat { p } .
4) Interpret the interval.
4
Why would manufacturers and businesses be interested in constructing a confidence interval for the population variance? Would manufacturers and businesses want large or small variances?
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5
When determining the sample size needed to achieve a particular error estimate you need to know ?. What are two methods of estimating ? if ? is unknown?
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6
 Why is s2 the best point estimate of σ2 ? \text { Why is } \mathrm { s } ^ { 2 } \text { the best point estimate of } \sigma ^ { 2 } \text { ? }
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7
Hannah selected a simple random sample of all adults in her town and, based on this sample, constructed a confidence interval for the mean salary of all adults in the town.
However, the distribution of salaries in the town is not exactly normal. Will the confidence interval still give a good estimate of the mean salary?
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8
How do you determine whether to use the z or t distribution in computing the margin of  error, E=zα/2σn or E=tα/2sn ? \text { error, } \mathrm { E } = \mathrm { z } _ { \alpha / 2 } \cdot \frac { \sigma } { \sqrt { \mathrm { n } } } \text { or } \mathrm { E } = \mathrm { t } _ { \alpha / 2 } \cdot \frac { \mathrm { s } } { \sqrt { \mathrm { n } } } \text { ? }
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9
Describe the steps for finding a confidence interval.
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10
Complete the table to compare z and t distributions. Complete the table to compare z and t distributions.
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11
Interpret the following 95% confidence interval for mean weekly salaries of shift managers at Guiseppe's Pizza and Pasta. 325.80 < µ < 472.30
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12
Under what circumstances can you replace ? with s in the formula E=zα/2σn\mathrm { E } = \mathrm { z } _ { \alpha / 2 } \cdot \frac { \sigma } { \sqrt { \mathrm { n } } }
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13
A researcher is interested in estimating the proportion of voters who favor a tax on e-commerce. Based on a sample of 250 people, she obtains the following 99% confidence interval for the population proportion, p:0.113<p<0.171\mathrm { p } : 0.113 < \mathrm { p } < 0.171 Which of the statements below is a valid interpretation
Of this confidence interval?

A) If 100 different samples of size 250 were selected and, based on each sample, a confidence interval were constructed, exactly 99 of these confidence intervals would contain the true value of p.
B) If many different samples of size 250 were selected and, based on each sample, a confidence interval were constructed, 99% of the time the true value of p would lie between 0.113 and 0.171.
C) There is a 99% chance that the true value of p lies between 0.113 and 0.171.
D) If many different samples of size 250 were selected and, based on each sample, a confidence interval were constructed, in the long run 99% of the confidence intervals would contain the true value of p.
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14
What assumption about the parent population is needed to use the t distribution to compute the margin of error?
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15
Based on a simple random sample of students from her school, Sally obtained a point estimate of the mean weight of students at her school. What additional information would be provided by a confidence interval estimate of the mean weight?
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16
What is the best point estimate for the population proportion? Explain why that point estimate is best.
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17
Explain how confidence intervals might be used to make decisions. Give an example to clarify your explanation.
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18
When determining sample size we need to know p^\hat { \mathrm { p } } . If we have no prior information, what are two methods that can be used?
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19
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20
Explain the difference between descriptive and inferential statistics.
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21
50 people are selected randomly from a certain population and it is found that 18 people in the sample are over 6 feet tall. What is the point estimate of the proportion of people in the population who are over 6 feet tall?

A) 0.50
B) 0.36
C) 0.64
D) 0.23
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22
A 99%99 \% confidence interval (in inches) for the mean height of a population is 66.2<μ<67.666.2 < \mu < 67.6 . This result is based on a sample of size 144 . Construct the 95%95 \% confidence interval. (Hint: you will first need to find the sample mean and sample standard deviation).

A) 66.266.2 in <μ<67.6< \mu < 67.6 in.
B) 66.066.0 in <μ<67.8< \mu < 67.8 in.
C) 66.366.3 in <μ<67.5< \mu < 67.5 in.
D) 66.466.4 in <μ<67.4in< \mu < 67.4 \mathrm { in } .
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23
Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.

-A sociologist develops a test to measure attitudes towards public transportation, and 27 randomly selected subjects are given the test. Their mean score is 76.2 and their standard deviation is 21.4.
Construct the 95% confidence interval for the mean score of all such subjects.

A) 74.6<μ<77.874.6 < \mu < 77.8
B) 67.7<μ<84.767.7 < \mu < 84.7
C) 64.2<μ<88.264.2 < \mu < 88.2
D) 69.2<μ<83.269.2 < \mu < 83.2
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24
Use the confidence level and sample data to find a confidence interval for estimating the population µ. Round your answer to the same number of decimal places as the sample mean.

-Test scores: n=79,x=65.0,σ=5.0;98%\mathrm { n } = 79 , \overline { \mathrm { x } } = 65.0 , \sigma = 5.0 ; 98 \% confidence

A) 64.1<μ<65.964.1 < \mu < 65.9
B) 63.9<μ<66.163.9 < \mu < 66.1
C) 63.5<μ<66.563.5 < \mu < 66.5
D) 63.7<μ<66.363.7 < \mu < 66.3
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25
The Bide-a-While efficiency hotel, which caters to business workers who stay for extended periods of time (weeks or months), offers room service. In a small study of 35 randomly selected room service orders, the 95% confidence interval for mean delivery time for room service is 24.8<μ<29.624.8 < \mu < 29.6 minutes. The marketing director is trying to determine if she can advertise "room service in under 30 minutes, or the order is free." How would you advise her?
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26
Use the confidence level and sample data to find a confidence interval for estimating the population µ. Round your answer to the same number of decimal places as the sample mean.

-A group of 64 randomly selected students have a mean score of 38.638.6 with a standard deviation of 4.94.9 on a placement test. What is the 90%90 \% confidence interval for the mean score, μ\mu , of all students taking the test?

A) 37.4<μ<39.837.4 < \mu < 39.8
B) 37.2<μ<40.037.2 < \mu < 40.0
C) 37.6<μ<39.637.6 < \mu < 39.6
D) 37.0<μ<40.237.0 < \mu < 40.2
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27
Under what three conditions is it appropriate to use the t distribution in place of the standard normal distribution?
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28
Use the given data to find the minimum sample size required to estimate the population proportion.

-Margin of error: 0.018; confidence level: 99%;p^ and q^ unknown 99 \% ; \hat { \mathrm { p } } \text { and } \hat { \mathrm { q } } \text { unknown }

A) 4966
B) 4114
C) 7116
D) 5117
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29
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.

- n=110,x=68;88%\mathrm { n } = 110 , \mathrm { x } = 68 ; 88 \% confidence

A) 0.545<p<0.6910.545 < p < 0.691
B) 0.541<p<0.6950.541 < p < 0.695
C) 0.542<p<0.6940.542 < p < 0.694
D) 0.546<p<0.6900.546 < p < 0.690
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30
In constructing a confidence interval for σ\sigma or σ2\sigma ^ { 2 } , a table is used to find the critical values χL2\chi _ { L } ^ { 2 } and χ2R\chi \underset { R } { 2 } for values of n101n \leq 101 . For larger values of n,χL2n , \chi _ { L } ^ { 2 } and χR2\chi { } _ { \mathrm { R } } ^ { 2 } can be approximated by using the following formula: χ2=12[±zα/2+2k1]2\chi ^ { 2 } = \frac { 1 } { 2 } [ \pm \mathrm { z } \alpha / 2 + \sqrt { 2 \mathrm { k } - 1 } ] ^ { 2 } where k\mathrm { k } is the number of degrees of freedom and zα/2\mathrm { z } \alpha / 2 is the critical z score. Construct the 90%90 \% confidence interval for σ\sigma using the following sample data: a sample of size n=234\mathrm { n } = 234 yields a mean weight of 155lb155 \mathrm { lb } and a standard deviation of 26.0lb26.0 \mathrm { lb } . Round the confidence interval limits to the nearest hundredth.

A) 24.13lb<σ<28.11lb24.13 \mathrm { lb } < \sigma < 28.11 \mathrm { lb }
B) 24.18lb<σ<28.18lb24.18 \mathrm { lb } < \sigma < 28.18 \mathrm { lb }
C) 24.57lb<σ<27.67lb24.57 \mathrm { lb } < \sigma < 27.67 \mathrm { lb }
D) 23.86lb<σ<28.63lb23.86 \mathrm { lb } < \sigma < 28.63 \mathrm { lb }
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31
A radio show host asked people to call in and say whether they support new legislation to promote cleaner sources of energy. Based on this sample, she constructed a confidence interval to estimate the proportion of all listeners to her show who support the legislation.
Is the confidence interval likely to give a good estimate of the proportion of her listeners who support the legislation?
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32
A paper published the results of a poll. It stated that, based on a sample of 1000 married men, 51% of married men say that they would marry the same woman again. The margin of error was given as ±3 percentage points and the confidence level was given as 95%.
What does it mean that the margin of error was ±3 percentage points?
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33
Use the given data to find the minimum sample size required to estimate the population proportion.

-Margin of error: 0.008; confidence level: 99%; from a prior study, p^\hat{p} is estimated by 0.208.

A) 9889
B) 15,361
C) 137
D) 17,068
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34
Define a point estimate. What is the best point estimate for Define a point estimate. What is the best point estimate for   ? ?
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35
Mark wanted to estimate the mean number of years of education of adults in his city. He waited outside a public library and interviewed every tenth adult leaving. Based on this sample, he constructed a confidence interval for the mean number of years of education of adults in the city. Do you think this confidence interval will give a good estimate? Why or why not?
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36
Draw a diagram of the chi-square distribution. Discuss its shape and values.
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37
Define confidence interval and degree of confidence. Make up an example of a confidence interval and interpret the result.
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38
Of 366 randomly selected medical students, 27 said that they planned to work in a rural community. Find a 95% confidence interval for the true proportion of all medical students who plan to work in a rural community.

A) 0.0513<p<0.09620.0513 < p < 0.0962
B) 0.0470<p<0.1010.0470 < p < 0.101
C) 0.0419<p<0.1060.0419 < p < 0.106
D) 0.0386<p<0.1090.0386 < p < 0.109
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39
Use the given data to find the minimum sample size required to estimate the population proportion.

-Margin of error: 0.0280.028 ; confidence level: 99%;p^99 \% ; \hat { \mathrm { p } } and q^\hat { \mathrm { q } } unknown

A) 2223
B) 1116
C) 2115
D) 1939
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40
Identify the correct distribution (z, t, or neither) for each of the following.  Sample Size  Standard  deviation  Shape of the  distribution  z or t or neither n=35s=4.5 Somewhat skewed n=20s=4.5 Bell shaped n=25σ=4.5 Bell shaped n=20σ=4.5 Extremely skewed \begin{array} { | l | l | l | l | } \hline \text { Sample Size } & \begin{array} { l } \text { Standard } \\\text { deviation }\end{array} & \begin{array} { l } \text { Shape of the } \\\text { distribution }\end{array} & \text { z or t or neither } \\\hline n = 35 & s = 4.5 & \text { Somewhat skewed } & \\\hline n = 20 & s = 4.5 & \text { Bell shaped } & \\\hline n = 25 & \sigma = 4.5 & \text { Bell shaped } & \\\hline n = 20 & \sigma = 4.5 & \text { Extremely skewed } & \\\hline\end{array}
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41
Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.

- 90% confidence; n=430,x=8090 \% \text { confidence; } n = 430 , x = 80

A) 0.0386
B) 0.0368
C) 0.0331
D) 0.0309
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42
A newspaper article about the results of a poll states: "In theory, the results of such a poll, in 99 cases out of 100 should differ by no more than 2 percentage points in either direction from what would have been obtained by interviewing all voters in the United States." Find the sample size suggested by this statement.

A) 2402
B) 4145
C) 165
D) 3394
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43
Do one of the following, as appropriate: (a) Find the critical value z?/2, (b) find the critical value t?/2, (c) state that neither the normal nor the t distribution applies.

- 90%;n=10;σ90 \% ; \mathrm { n } = 10 ; \sigma is unknown; population appears to be normally distributed.

A) tα/2=1.833\mathrm { t } _ { \alpha / 2 } = 1.833
B) zα/2=2.262\mathrm { z } _ { \alpha / 2 } = 2.262
C) tα/2=1.812\mathrm { t } _ { \alpha / 2 } = 1.812
D) zα/2=1.383\mathrm { z } _ { \alpha / 2 } = 1.383
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44
In a clinical test with 8900 subjects, 4450 showed improvement from the treatment. Find the margin of error for the 99% confidence interval used to estimate the population proportion.

A) 0.00780
B) 0.0137
C) 0.0120
D) 0.0104
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45
Use the given information to find the minimum sample size required to estimate an unknown population mean µ.

-  Margin of error: $134, confidence level: 95%,σ=$575\text { Margin of error: } \$ 134 \text {, confidence level: } 95 \% , \sigma = \$ 575

A) 71
B) 62
C) 50
D) 100
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46
Use the given data to find the minimum sample size required to estimate the population proportion.

-  Margin of error: 0.008; confidence level: 99%;p^ and q^ unknown \text { Margin of error: } 0.008 \text {; confidence level: } 99 \% ; \hat { \mathrm { p } } \text { and } \hat { \mathrm { q } } \text { unknown }

A) 25,901
B) 15,900
C) 26,024
D) 25,894
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47
Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation ?.
Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation.

-Weights of men: 90%90 \% confidence; n=14,x=154.8lb,s=11.1lb\mathrm { n } = 14 , \overline { \mathrm { x } } = 154.8 \mathrm { lb } , \mathrm { s } = 11.1 \mathrm { lb }

A) 8.7lb<σ<14.3lb8.7 \mathrm { lb } < \sigma < 14.3 \mathrm { lb }
B) 8.2lb<σ<15.6lb8.2 \mathrm { lb } < \sigma < 15.6 \mathrm { lb }
C) 9.0lb<σ<2.7lb9.0 \mathrm { lb } < \sigma < 2.7 \mathrm { lb }
D) 8.5lb<σ<16.5lb8.5 \mathrm { lb } < \sigma < 16.5 \mathrm { lb }
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48
Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.

-The principal randomly selected six students to take an aptitude test. Their scores were: 81.478.671.377.876.977.7\begin{array} { l l l l l l } 81.4 & 78.6 & 71.3 & 77.8 & 76.9 & 77.7 \end{array}
Determine a 90%90 \% confidence interval for the mean score for all students.

A) 80.11<μ<74.4580.11 < \mu < 74.45
B) 80.01<μ<74.5580.01 < \mu < 74.55
C) 74.55<μ<80.0174.55 < \mu < 80.01
D) 74.45<μ<80.1174.45 < \mu < 80.11
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49
Solve the problem. Round the point estimate to the nearest thousandth.
364 randomly selected light bulbs were tested in a laboratory, 124 lasted more than 500 hours. Find a point estimate of the proportion of all light bulbs that last more than 500 hours.

A) 0.254
B) 0.659
C) 0.338
D) 0.341
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50
Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation ?.
Assume that the population has a normal distribution. Round the confidence interval limits to one more decimal place than is used for the original set of data.

-The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were:
7913675951513\begin{array} { c c c c c } 7 & 9 & 13 & 6 & 7 \\5 & 9 & 5 & 15 & 13\end{array}
Find a 95%95 \% confidence interval for the population standard deviation σ\sigma .

A) 0.8 min<σ<2.4 min0.8 \mathrm {~min} < \sigma < 2.4 \mathrm {~min}
B) 2.4 min<σ<6.0 min2.4 \mathrm {~min} < \sigma < 6.0 \mathrm {~min}
C) 2.5 min<σ<6.0 min2.5 \mathrm {~min} < \sigma < 6.0 \mathrm {~min}
D) 2.5 min<σ<6.6 min2.5 \mathrm {~min} < \sigma < 6.6 \mathrm {~min}
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51
Use the given data to find the minimum sample size required to estimate the population proportion.

-Margin of error: 0.090.09 ; confidence level: 95%95 \% ; from a prior study, p^\hat { \mathrm { p } } is estimated by the decima equivalent of 87%87 \% .

A) 5
B) 48
C) 54
D) 162
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52
Find the critical value χR2\chi { } _ { \mathrm { R } } ^ { 2 } corresponding to a sample size of 19 and a confidence level of 99 percent.

A) 34.80534.805
B) 37.15637.156
C) 6.2656.265
D) 7.0157.015
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53
Use the given information to find the minimum sample size required to estimate an unknown population mean µ.

-How many students must be randomly selected to estimate the mean weekly earnings of students at one college? We want 95% confidence that the sample mean is within $2 of the population mean, and the population standard deviation is known to be $10.

A) 136
B) 97
C) 68
D) 85
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54
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.

- n=150,x=54;90%\mathrm { n } = 150 , \mathrm { x } = 54 ; 90 \% confidence

A) 0.299<p<0.4210.299 < p < 0.421
B) 0.296<p<0.4240.296 < p < 0.424
C) 0.298<p<0.4220.298 < p < 0.422
D) 0.294<p<0.4260.294 < p < 0.426
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55
Use the given information to find the minimum sample size required to estimate an unknown population mean µ.

-  Margin of error: $136, confidence level: 99%,σ=$584\text { Margin of error: } \$ 136 \text {, confidence level: } 99 \% , \sigma = \$ 584

A) 62
B) 123
C) 50
D) 71
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56
Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.

-A savings and loan association needs information concerning the checking account balances of its local customers. A random sample of 14 accounts was checked and yielded a mean balance of $664.14 and a standard deviation of $297.29. Find a 98% confidence interval for the true mean checking account balance for local customers.

A) $492.52<μ<$835.76\$ 492.52 < \mu < \$ 835.76
B) $493.71<μ<$834.57\$ 493.71 < \mu < \$ 834.57
C) $453.59<μ<$874.69\$ 453.59 < \mu < \$ 874.69
D) $455.65<μ<$872.63\$ 455.65 < \mu < \$ 872.63
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57
Use the given data to find the minimum sample size required to estimate the population proportion.

-  Margin of error: 0.009; confidence level: 99%;p^ and q^ unknown \text { Margin of error: 0.009; confidence level: } 99 \% ; \hat { \mathrm { p } } \text { and } \hat { \mathrm { q } } \text { unknown }

A) 20,465
B) 21,442
C) 9642
D) 19,566
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58
Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.

-90% confidence; the sample size is 1410, of which 40% are successes

A) 0.0167
B) 0.0267
C) 0.0256
D) 0.0215
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59
Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.

-95% confidence; the sample size is 6100, of which 40% are successes

A) 0.00923
B) 0.0141
C) 0.0123
D) 0.0162
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60
Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.

- 95% confidence; n=470,x=5095 \% \text { confidence; } \mathrm { n } = 470 , \mathrm { x } = 50

A) 0.0251
B) 0.0335
C) 0.0293
D) 0.0279
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61
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.

-Of 80 adults selected randomly from one town, 67 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance.

A) 0.741<p<0.9340.741 < \mathrm { p } < 0.934
B) 0.731<p<0.9440.731 < \mathrm { p } < 0.944
C) 0.757<p<0.9180.757 < p < 0.918
D) 0.770<p<0.9050.770 < p < 0.905
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62
Use the confidence level and sample data to find a confidence interval for estimating the population µ. Round your answer to the same number of decimal places as the sample mean.

-A random sample of 108 light bulbs had a mean life of xˉ=547\bar { x } = 547 hours with a standard deviation of σ=36\sigma = 36 hours. Construct a 90%90 \% confidence interval for the mean life, μ\mu , of all light bulbs of this type.

A) 541hr<μ<553hr541 \mathrm { hr } < \mu < 553 \mathrm { hr }
B) 538hr<μ<556hr538 \mathrm { hr } < \mu < 556 \mathrm { hr }
C) 539hr<μ<555hr539 \mathrm { hr } < \mu < 555 \mathrm { hr }
D) 540hr<μ<554hr540 \mathrm { hr } < \mu < 554 \mathrm { hr }
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63
Use the given information to find the minimum sample size required to estimate an unknown population mean µ.

-How many women must be randomly selected to estimate the mean weight of women in one age group. We want 90% confidence that the sample mean is within 2.8 lb of the population mean, and the population standard deviation is known to be 27 lb.

A) 358
B) 253
C) 252
D) 250
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64
Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.

-Thirty randomly selected students took the calculus final. If the sample mean was 75 and the standard deviation was 13.2, construct a 99% confidence interval for the mean score of all students.

A) 68.38<μ<81.6268.38 < \mu < 81.62
B) 70.91<μ<79.0970.91 < \mu < 79.09
C) 68.36<μ<81.6468.36 < \mu < 81.64
D) 69.07<μ<80.9369.07 < \mu < 80.93
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65
The amounts (in ounces) of juice in eight randomly selected juice bottles are:
15.315.315.715.715.315.915.315.9\begin{array} { l l l l l } 15.3 & 15.3 & 15.7 & 15.7\\15.3&15 .9&15 .3&15 .9\\ \end{array}
Construct a 98%98 \% confidence interval for the mean amount of juice in all such bottle

A) 15.83oz<μ<15.27oz15.83 \mathrm { oz } < \mu < 15.27 \mathrm { oz }
B) 15.93oz<μ<15.17oz15.93 \mathrm { oz } < \mu < 15.17 \mathrm { oz }
C) 15.27oz<μ<15.83oz15.27 \mathrm { oz } < \mu < 15.83 \mathrm { oz }
D) 15.17oz<μ<15.93oz15.17 \mathrm { oz } < \mu < 15.93 \mathrm { oz }
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66
Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation ?.
Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation.

-A sociologist develops a test to measure attitudes about public transportation, and 27 randomly selected subjects are given the test. Their mean score is 76.2 and their standard deviation is 21.4.
Construct the 95% confidence interval for the standard deviation, ?, of the scores of all subjects.

A) 16.9<σ<29.316.9 < \sigma < 29.3
B) 17.5<σ<27.817.5 < \sigma < 27.8
C) 16.6<σ<28.616.6 < \sigma < 28.6
D) 17.2<σ<27.217.2 < \sigma < 27.2
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67
Use the given data to find the minimum sample size required to estimate the population proportion.

-Margin of error: 0.04; confidence level: 99%; from a prior study, p^\hat { p } is estimated by 0.14.

A) 499
B) 20
C) 289
D) 599
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68
Express the confidence interval using the indicated format.

-Express the confidence interval 0.307±0.0580.307 \pm 0.058 in the form of p^E<p<p^+E\hat { p } - \mathrm { E } < \mathrm { p } < \hat { \mathrm { p } } + \mathrm { E } .

A) 0.278<p<0.3360.278 < p < 0.336
B) 0.307<p<0.3650.307 < p < 0.365
C) 0.249<p<0.3070.249 < p < 0.307
D) 0.249<p<0.3650.249 < p < 0.365
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69
Find the critical value zα/2\mathrm { z } _ { \alpha / 2 } that corresponds to a 98% confidence level.

A) 2.575
B) 2.33
C) 2.05
D) 1.75
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70
Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.

-99% confidence; the sample size is 1180, of which 45% are successes

A) 0.0297
B) 0.0284
C) 0.0337
D) 0.0373
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71
Find the critical value χR2\chi _ { \mathrm { R } } ^ { 2 } corresponding to a sample size of 7 and a confidence level of 90 percent.

A) 1.635
B) 12.592
C) 18.548
D) 16.812
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72
Express the confidence interval using the indicated format.

-Express the confidence interval 0.62<p<0.720.62 < p < 0.72 in the form of p^±E\hat { p } \pm E .

A) 0.67±0.050.67 \pm 0.05
B) 0.62±0.10.62 \pm 0.1
C) 0.62±0.050.62 \pm 0.05
D) 0.67±0.10.67 \pm 0.1
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73
 Find zα/2 for α=0.06\text { Find } z _ { \alpha / 2 } \text { for } \alpha = 0.06

A) 1.88
B) 1.555
C) 1.96
D) 2.75
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74
Use the confidence level and sample data to find a confidence interval for estimating the population µ. Round your answer to the same number of decimal places as the sample mean.

-Test scores: n=102,x=77.4,σ=6.5;99%\mathrm { n } = 102 , \overline { \mathrm { x } } = 77.4 , \sigma = 6.5 ; 99 \% confidence

A) 76.1<μ<78.776.1 < \mu < 78.7
B) 75.7<μ<79.175.7 < \mu < 79.1
C) 75.9<μ<78.975.9 < \mu < 78.9
D) 76.3<μ<78.576.3 < \mu < 78.5
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75
The following confidence interval is obtained for a population proportion, p:0.686<p<0.712p : 0.686 < p < 0.712 . Use these confidence interval limits to find the point estimate, p^\hat {\mathrm { p }} .

A) 0.699
B) 0.704
C) 0.686
D) 0.694
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76
Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.

- n=10,x=12.7, s=3.7,95%\mathrm { n } = 10 , \overline { \mathrm { x } } = 12.7 , \mathrm {~s} = 3.7,95 \% confidence

A) 10.07<μ<15.3310.07 < \mu < 15.33
B) 10.05<μ<15.3510.05 < \mu < 15.35
C) 10.09<μ<15.3110.09 < \mu < 15.31
D) 10.56<μ<14.8410.56 < \mu < 14.84
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77
n=12,x=21.9, s=4.0,99%\mathrm { n } = 12 , \overline { \mathrm { x } } = 21.9 , \mathrm {~s} = 4.0,99 \% confidence

A) 18.24<μ<25.5618.24 < \mu < 25.56
B) 18.76<μ<25.0418.76 < \mu < 25.04
C) 18.31<μ<25.4918.31 < \mu < 25.49
D) 18.33<μ<25.4718.33 < \mu < 25.47
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78
Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.

-A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 198 milligrams with s=10.5\mathrm { s } = 10.5 milligrams. Construct a 95%95 \% confidence interval for the true mean cholesterol content of all such eggs.

A) 191.3mg<μ<204.7mg191.3 \mathrm { mg } < \mu < 204.7 \mathrm { mg }
B) 191.4mg<μ<204.6mg191.4 \mathrm { mg } < \mu < 204.6 \mathrm { mg }
C) 191.2mg<μ<204.8mg191.2 \mathrm { mg } < \mu < 204.8 \mathrm { mg }
D) 192.6mg<μ<203.4mg192.6 \mathrm { mg } < \mu < 203.4 \mathrm { mg }
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79
Use the given data to find the minimum sample size required to estimate the population proportion.

-Margin of error: 0.008; confidence level 95%;p^ and q^ unknown 95 \% ; \hat { \mathrm { p } } \text { and } \hat { \mathrm { q } } \text { unknown }

A) 15,007
B) 15,098
C) 14,488
D) 5045
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80
Find the chi-square value xL2x _ { L } ^ { 2 } corresponding to a sample size of 4 and a confidence level of 98 percent.

A) 0.216
B) 9.348
C) 0.115
D) 11.345
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