Question 1

The director of a hospital wishes to estimate the mean number of people who are admitted to the emergency room during a 24-hour period. The director randomly selects 64 different 24-hour
Periods and determines the number of admissions for each. For this sample, x = 19.8 and s2 = 36. If
The director wishes to estimate the mean number of admissions per 24-hour period to within 1
Admission with 90% reliability, what is the minimum sample size she should use?
A)3508
B)60
C)98
D)2132

Accepted Answer

The minimum sample size can be calculated using the formula for the sample size of a population mean: $n = \left(\frac{Z \cdot \sigma}{E}\right)^2$, where $Z$ is the Z-score corresponding to the desired confidence level (90% in this case, which gives a Z-score of approximately 1.645), $\sigma$ is the standard deviation (the square root of $s^2$, which is 6), and $E$ is the margin of error (1 admission). Plugging in the values: $n = \left(\frac{1.645 \cdot 6}{1}\right)^2 = \left(9.87\right)^2 \approx 97.4$, which rounds up to 98 for a whole number of samples.

Question 2

It is desired to estimate the proportion of college students who feel a sudden relief now that their statistics class is over. How many students must be sampled in order to estimate the true
Proportion to within 2% at the 90% confidence level?
A)n = 189
B)Cannot determine because no estimate of p or q exists in this problem
C)n = 2401
D)n = 1692
E)n = 133

Accepted Answer

The answer of It is desired to estimate the proportion...

Question 3

A local menʹs clothing store is being sold. The buyers are trying to estimate the percentage
of items that are outdated. They will choose a random sample from the 100,000 items in
the storeʹs inventory in order to determine the proportion of merchandise that is outdated.
The current owners have never determined the percentage of outdated merchandise and
cannot help the buyers. How large a sample do the buyers need in order to be 90%
confident that the margin of error of their estimate is within 3%?

Accepted Answer

The answer of A local menʹs clothing store is being...

Question 4

A previous random sample of 4000 U.S. citizens yielded 2250 who are in favor of gun control legislation. How many citizens would need to be sampled for a 99% confidence interval to estimate
The true proportion within 3%?
A)1814
B)1695
C)1916
D)1842

Accepted Answer

The sample size needed for a 99% confidence interval with a margin of error of 3% is 1814. This can be calculated using the formula: n = (Z^2 * p * q) / e^2, where Z is the z-score for the desired confidence level (2.576 for 99%), p is the sample proportion (0.5625), q is the complement of p (1 - p = 0.4375), and e is the margin of error (0.03).

Question 5

A confidence interval was used to estimate the proportion of statistics students who are female. A random sample of 72 statistics students generated the following confidence interval: (.438, .642).
Using the information above, what sample size would be necessary if we wanted to estimate the
True proportion to within 3% using 99% reliability?
A)1916
B)1831
C)1769
D)1842

Accepted Answer

The answer of A confidence interval was used to estimate...

Question 6

The standard deviation of a population is estimated to be 345 units. To estimate the
population mean to within 49 units with 99% reliability, what size sample should be
selected?

Accepted Answer

The answer of The standard deviation of a population is...

Question 7

Suppose it is desired to estimate the average time a customer spends in a particular store to within 5 minutes $$( \text { e.g. } \pm 5 \text { minutes) }$$ at 99% reliability. It is estimated that the standard deviation of the
Times is 15 minutes. How large a sample should be taken to get the desired interval?
A)n = 299
B)n = 60
C)n = 25
D)n = 35

Accepted Answer

The answer of Suppose it is desired to estimate the...

Question 8

It is desired to estimate the average time it takes Statistics students to finish a computer project to within two hours at 90% reliability. It is estimated that the standard deviation of the times is 14
Hours. How large a sample should be taken to get the desired interval?
A)n = 133
B)n = 325
C)n = 189
D)n = 231

Accepted Answer

The sample size needed to estimate the average time it takes Statistics students to finish a computer project to within two hours at 90% reliability is:n = (Z^2 * s^2) / E^2= (1.645^2 * 14^2) / 2^2= 133.12Therefore, the correct answer is A) n = 133.

Question 9

Suppose you wanted to estimate a binomial proportion, p, correct to within .05 with
probability 0.95. What size sample would need to be selected if p is known to be
approximately 0.8?

Accepted Answer

The answer of Suppose you wanted to estimate a binomial...

Question 10

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200
Students and finds that 118 of them are receiving financial aid. If the dean wanted to estimate the
Proportion of all students receiving financial aid to within 1% with 99% reliability, how many
Students would need to be sampled?
A)16,040
B)6229
C)161
D)3880

Accepted Answer

To estimate the proportion with 1% margin of error and 99% confidence, the sample size formula n = (Z^2 * p * (1-p)) / E^2 is used. With Z=2.576 for 99% confidence, p=0.59 (from 118/200), and E=0.01, the calculation gives approximately 16,040.

Question 11

In the construction of confidence intervals, if all other quantities are unchanged, an increase in the sample size will lead to a __________ interval.
A)biased
B)wider
C)less significant
D)narrower

Accepted Answer

An increase in sample size generally leads to a narrower confidence interval because it reduces the standard error, making the estimate more precise.

Question 12

Sales of a new line of athletic footwear are crucial to the success of a company. The company wishes to estimate the average weekly sales of the new footwear to within $200 with 95%
Reliability. The initial sales indicate that the standard deviation of the weekly sales figures is
Approximately $1400. How many weeks of data must be sampled for the company to get the
Information it desires?
A)14 weeks
B)97 weeks
C)189 weeks
D)37,648 weeks

Accepted Answer

To find the sample size (n), use the formula for the sample size of a mean: n = (Z*σ/E)^2. Here, Z is the Z-score for 95% confidence (1.96), σ is the standard deviation ($1400), and E is the margin of error ($200). Plugging in these values gives n = (1.96*1400/200)^2 = 189 weeks.

Question 13

For the given combination of $\alpha$ and degrees of freedom (df), find the value of $\chi _ { ( 1 - \alpha / 2 ) } ^ { 2 }$ that would be used to find the upper endpoint of a confidence interval for $\sigma ^ { 2 }$. $\alpha = 0.01 , \mathrm { df } = 30$
A) $13.7867$
B) $14.9535$
C) $53.6720$
D) $13.1211$

Accepted Answer

The correct answer is A because the value of $\chi _ { ( 1 - \alpha / 2 ) } ^ { 2 }$ that would be used to find the upper endpoint of a confidence interval for $\sigma ^ { 2 }$ is 13.7867. This can be found using a chi-square distribution table or a calculator.

Question 14

Given the values of $\bar { x } , \mathrm {~s}$, and $n$, form a $99 \%$ confidence interval for $\sigma$. $\overline { \mathrm { x } } = 19.7 , \mathrm {~s} = 4.8 , \mathrm { n } = 14$
A) $( 3.17,9.17 )$
B) $( 10.04,84.02 )$
C) $( 3.29,8.54 )$
D) $( 2.09,17.5 )$

Accepted Answer

The answer of Given the values of \(\bar { x...

Question 15

If no estimate of p exists when determining the sample size for a confidence interval for a
proportion, we can use .5 in the formula to get a value for n.

Accepted Answer

Using 0.5 for p in the formula to determine sample size for a confidence interval for a proportion maximizes the sample size estimate, ensuring the sample is large enough for a wide range of possible true proportions.

Question 16

Given the values of $\bar { x } , \mathrm {~s}$, and $\mathrm { n }$, form a $99 \%$ confidence interval for $\sigma ^ { 2 }$. $\overline { \mathrm { x } } = 15.1 , \mathrm {~s} = 4.3 , \mathrm { n } = 24$
A) $( 10.04,47.92 )$
B) $( 9.63,45.92 )$
C) $( 10.21,41.71 )$
D) $( 2.38,9.7 )$

Accepted Answer

The confidence interval for $\sigma^2$ is calculated using the chi-square distribution with $(n-1)$ degrees of freedom. The correct interval is $( 9.63,45.92 )$.

Question 17

After elections were held, it was desired to estimate the proportion of voters who regretted that they did not vote. How many voters must be sampled in order to estimate the true proportion to within $2 \% ($ e.g., $\pm 0.02$ ) at the $90 \%$ confidence level? Assume that we believe this proportion lies close to $30 \%$.
A) $n = 2401$
B) $n = 1421$
C) $\mathrm { n } = 1692$
D) $n = 2017$
E) Cannot determine because no estimate of $\mathrm { p }$ or $\mathrm { q }$ exists in this problem.

Accepted Answer

To estimate the sample size needed for a proportion with a specific margin of error and confidence level, the formula is $n = \left(\frac{Z^2 \cdot p \cdot (1-p)}{E^2}\right)$, where $Z$ is the Z-score corresponding to the confidence level (for 90%, $Z \approx 1.645$), $p$ is the estimated proportion (0.30 in this case), and $E$ is the margin of error (0.02). Plugging in the values: $n = \left(\frac{1.645^2 \cdot 0.30 \cdot (1-0.30)}{0.02^2}\right) \approx 1421$.

Question 18

Sales of a new line of athletic footwear are crucial to the success of a newly formed
company. The company wishes to estimate the average weekly sales of the new footwear
to within $150 with 99% reliability. The initial sales indicate that the standard deviation of
the weekly sales figures is approximately $1550. How many weeks of data must be
sampled for the company to get the information it desires?

Accepted Answer

The answer of Sales of a new line of athletic...

Question 19

One way of reducing the width of a confidence interval is to reduce the size of the sample taken.

Accepted Answer

Reducing the size of the sample typically increases the width of the confidence interval because a smaller sample size increases the uncertainty (or standard error) of the estimate.

Question 20

For the given combination of $\alpha$ and degrees of freedom (df), find the value of $\chi _ { \alpha / 2 } ^ { 2 }$ that would be used to find the lower endpoint of a confidence interval for $\sigma ^ { 2 }$. $\alpha = 0.1 , \mathrm { df } = 14$
A) $6.5706$
B) $23.6848$
C) $21.0642$
D) $22.3621$

Accepted Answer

The answer of For the given combination of $\alpha$ and...

Quiz 7: Inferences Based on a Single Sample: Estimation With Confidence Intervals

It is desired to estimate the proportion of college students who feel a sudden relief now that their statistics class is over. How many students must be sampled in order to estimate the true Proportion to within 2% at the 90% confidence level?

A previous random sample of 4000 U.S. citizens yielded 2250 who are in favor of gun control legislation. How many citizens would need to be sampled for a 99% confidence interval to estimate The true proportion within 3%?

The standard deviation of a population is estimated to be 345 units. To estimate the population mean to within 49 units with 99% reliability, what size sample should be selected?

It is desired to estimate the average time it takes Statistics students to finish a computer project to within two hours at 90% reliability. It is estimated that the standard deviation of the times is 14 Hours. How large a sample should be taken to get the desired interval?

Suppose you wanted to estimate a binomial proportion, p, correct to within .05 with probability 0.95. What size sample would need to be selected if p is known to be approximately 0.8?

In the construction of confidence intervals, if all other quantities are unchanged, an increase in the sample size will lead to a __________ interval.

For the given combination of $\alpha$ and degrees of freedom (df) , find the value of $\chi _ { ( 1 - \alpha / 2 ) } ^ { 2 }$ that would be used to find the upper endpoint of a confidence interval for $\sigma ^ { 2 }$ . $\alpha = 0.01 , \mathrm { df } = 30$

Given the values of $\bar { x } , \mathrm {~s}$ , and $n$ , form a $99 \%$ confidence interval for $\sigma$ . $\overline { \mathrm { x } } = 19.7 , \mathrm {~s} = 4.8 , \mathrm { n } = 14$

If no estimate of p exists when determining the sample size for a confidence interval for a proportion, we can use .5 in the formula to get a value for n.

Given the values of $\bar { x } , \mathrm {~s}$ , and $\mathrm { n }$ , form a $99 \%$ confidence interval for $\sigma ^ { 2 }$ . $\overline { \mathrm { x } } = 15.1 , \mathrm {~s} = 4.3 , \mathrm { n } = 24$

One way of reducing the width of a confidence interval is to reduce the size of the sample taken.

For the given combination of $\alpha$ and degrees of freedom (df) , find the value of $\chi _ { \alpha / 2 } ^ { 2 }$ that would be used to find the lower endpoint of a confidence interval for $\sigma ^ { 2 }$ . $\alpha = 0.1 , \mathrm { df } = 14$

Quiz 7: Inferences Based on a Single Sample: Estimation With Confidence Intervals

It is desired to estimate the proportion of college students who feel a sudden relief now that their statistics class is over. How many students must be sampled in order to estimate the true Proportion to within 2% at the 90% confidence level?

A previous random sample of 4000 U.S. citizens yielded 2250 who are in favor of gun control legislation. How many citizens would need to be sampled for a 99% confidence interval to estimate The true proportion within 3%?

The standard deviation of a population is estimated to be 345 units. To estimate the population mean to within 49 units with 99% reliability, what size sample should be selected?

It is desired to estimate the average time it takes Statistics students to finish a computer project to within two hours at 90% reliability. It is estimated that the standard deviation of the times is 14 Hours. How large a sample should be taken to get the desired interval?

Suppose you wanted to estimate a binomial proportion, p, correct to within .05 with probability 0.95. What size sample would need to be selected if p is known to be approximately 0.8?

In the construction of confidence intervals, if all other quantities are unchanged, an increase in the sample size will lead to a __________ interval.

Given the values of xˉ, s\bar { x } , \mathrm {~s}xˉ, s , and nnn , form a 99%99 \%99% confidence interval for σ\sigmaσ . x‾=19.7, s=4.8,n=14\overline { \mathrm { x } } = 19.7 , \mathrm {~s} = 4.8 , \mathrm { n } = 14x=19.7, s=4.8,n=14

If no estimate of p exists when determining the sample size for a confidence interval for a proportion, we can use .5 in the formula to get a value for n.

Given the values of xˉ, s\bar { x } , \mathrm {~s}xˉ, s , and n\mathrm { n }n , form a 99%99 \%99% confidence interval for σ2\sigma ^ { 2 }σ2 . x‾=15.1, s=4.3,n=24\overline { \mathrm { x } } = 15.1 , \mathrm {~s} = 4.3 , \mathrm { n } = 24x=15.1, s=4.3,n=24

One way of reducing the width of a confidence interval is to reduce the size of the sample taken.

For the given combination of α\alphaα and degrees of freedom (df) , find the value of χα/22\chi _ { \alpha / 2 } ^ { 2 }χα/22​ that would be used to find the lower endpoint of a confidence interval for σ2\sigma ^ { 2 }σ2 . α=0.1,df=14\alpha = 0.1 , \mathrm { df } = 14α=0.1,df=14

Given the values of $\bar { x } , \mathrm {~s}$ , and $n$ , form a $99 \%$ confidence interval for $\sigma$ . $\overline { \mathrm { x } } = 19.7 , \mathrm {~s} = 4.8 , \mathrm { n } = 14$

Given the values of $\bar { x } , \mathrm {~s}$ , and $\mathrm { n }$ , form a $99 \%$ confidence interval for $\sigma ^ { 2 }$ . $\overline { \mathrm { x } } = 15.1 , \mathrm {~s} = 4.3 , \mathrm { n } = 24$

For the given combination of $\alpha$ and degrees of freedom (df) , find the value of $\chi _ { \alpha / 2 } ^ { 2 }$ that would be used to find the lower endpoint of a confidence interval for $\sigma ^ { 2 }$ . $\alpha = 0.1 , \mathrm { df } = 14$