Question 1

Find the confidence interval specified.
-49 packages are randomly selected from packages received by a parcel service. The sample has a mean weight of 11.7 pounds. Assume that $\sigma$ = 3.1 pounds. What is the 95% confidence interval for
The true mean weight, µ, of all packages received by the parcel service?

Accepted Answer

The confidence interval is calculated using the formula: mean ± (Z * (σ/√n)). For a 95% confidence interval, Z is approximately 1.96. Plugging in the values: 11.7 ± (1.96 * (3.1/√49)) gives the interval 10.8 to 12.6 pounds.

Question 2

Solve the problem.
-A sample of 55 eggs yields a mean weight of 1.76 ounces. Assuming that $\sigma$ = 0.28 ounces, find the margin of error in estimating µ at the 95% level of confidence.

Accepted Answer

The margin of error (E) is calculated using the formula $E = z \times \frac{\sigma}{\sqrt{n}}$, where $z$ is the z-score for the 95% confidence level (approximately 1.96), $\sigma$ is the standard deviation, and $n$ is the sample size. Plugging in the values: $E = 1.96 \times \frac{0.28}{\sqrt{55}} \approx 0.07$ ounces.

Question 3

Find the specified t-value.
-For a t-curve with $d f = 20$, find $t _ { 0.01 }$.

Accepted Answer

For a t-curve with 20 degrees of freedom, $t_{0.01}$ refers to the t-value with a right-tail probability of 0.01. This value can be found in a t-distribution table or using statistical software. The correct value is $2.528$, which corresponds to the t-value for a 99% confidence level (or 0.01 significance level) with 20 degrees of freedom.

Question 4

Find the confidence interval specified.
-A sample of 38 people were randomly selected from among the workers in a shoe factory. The time taken for each person to polish a finished shoe was measured. The sample mean was 4.0 minutes.
Assume that $\sigma =$ 0.71 minutes. Construct a 90% confidence interval for the true mean time, $\mu$ to Polish a shoe.

Accepted Answer

The confidence interval is calculated using the formula for a Z-interval since the population standard deviation ($\sigma$) is known. The formula is $\bar{x} \pm Z\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $Z$ is the Z-score corresponding to the desired confidence level (for 90% confidence, $Z \approx 1.645$), $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values ($\bar{x} = 4.0$, $\sigma = 0.71$, $n = 38$), we find the margin of error and then the confidence interval, which matches choice B.

Question 5

Suppose that you wish to obtain a confidence interval for a population mean. Under the conditions described below, should you use the z-interval procedure, the t-interval procedure, or neither?
- The population standard deviation is known.
- The population is not normally distributed.
- The sample size is 12.

Accepted Answer

The z-interval procedure requires a known population standard deviation and either a large sample size (usually n ≥ 30) or a normally distributed population. The t-interval procedure is used when the population standard deviation is unknown, regardless of the sample size if the population is normally distributed, or if the sample size is large enough for the Central Limit Theorem to apply. Since the population standard deviation is known but the population is not normally distributed and the sample size is small (n = 12), neither procedure is appropriate.

Question 6

Solve the problem.
-Based on a sample of 36 randomly selected years, a 90% confidence interval for the mean annual precipitation in one city is from 46.8 inches to 49.2 inches. Find the margin of error.

Accepted Answer

The margin of error is half the width of the confidence interval, which is (49.2 - 46.8) / 2 = 1.2 inches.

Question 7

Find the necessary sample size.
-The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business
Students. Find the sample size needed to have a confidence level of 95% and a margin of error of
128 dollars.

Accepted Answer

To find the sample size, use the formula for the sample size of a mean: n = (Z*σ/E)^2. For a 95% confidence level, Z is approximately 1.96. Plugging in the values: n = (1.96*545/128)^2 ≈ 70.

Question 8

Find the specified t-value.
-For a t-curve with $\mathrm { df } = 24$, find $\mathrm { t } _ { 0.005 }$. $^ { \text {. } }$

Accepted Answer

The t-value $\mathrm { t } _ { 0.005 }$ for a t-distribution with 24 degrees of freedom is approximately 2.797. This value is found using a t-distribution table or calculator for the 0.005 significance level (two-tailed).

Question 9

Find the specified t-value.
-For a t-curve with df = 6, find the two t-values that divide the area under the curve into a middle 0.99 area and two outside areas of

Accepted Answer

For a t-distribution with 6 degrees of freedom (df = 6) and a middle area of 0.99, the two t-values that divide the curve into the specified areas are found using t-distribution tables or software for the 0.005 (0.5% in each tail since 1 - 0.99 = 0.01 and 0.01/2 = 0.005) significance level. The correct values are -3.707 and 3.707, which correspond to the critical t-values for a two-tailed test with a total area of 1% outside the middle 99% area.

Question 10

Solve the problem.
-A confidence interval for a population mean has a margin of error of 6.4. Determine the length of the confidence interval.

Accepted Answer

The length of the confidence interval is twice the margin of error, so it's 2 * 6.4 = 12.8.

Question 11

Find the confidence interval specified. Assume that the population is normally distributed.
-The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes)were: $$\begin{array} { r r r r r } 
12.9 & 13.3 & 13.3 & 7.7 & 11.7 \
9.9 & 12.5 & 6.5 & 10.8 & 12.2
\end{array}$$ Determine a 95% confidence interval for the mean time for all players.

Accepted Answer

The sample mean and standard deviation are calculated from the given data. Using the t-distribution for a 95% confidence interval with 9 degrees of freedom, the interval is calculated to be approximately 9.40 to 12.80 minutes.

Question 12

Find the confidence interval specified. Assume that the population is normally distributed.
-A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 188 milligrams with s = 12.7 milligrams. Construct a 95% confidence interval for the true mean
Cholesterol content of all such eggs.

Accepted Answer

The confidence interval is calculated using the formula: mean ± (t-value * (s/√n)). With n=12, the t-value for 95% confidence is approximately 2.201. The interval is 188 ± (2.201 * (12.7/√12)), which gives 179.9 to 196.1 milligrams.

Question 13

Solve the problem.
-A confidence interval for a population mean, µ, has l $\mu _ { ,}$ ength 179.1. Find the margin of error.

Accepted Answer

The margin of error is half the length of the confidence interval, so it is $179.1 / 2 = 89.55$.

Question 14

Provide an appropriate response.
-In which of the following situations is it reasonable to use the z-interval procedure to obtain a confidence interval for the population mean? Assume that the population standard deviation is
Known. A. $\mathrm { n } = 10$, the data contain no outliers, the variable under consideration is not normally distributed.
B. $\mathrm { n } = 10$, the variable under consideration is normally distributed.
C. $\mathrm { n } = 18$, the data contain no outliers, the variable under consideration is far from being normally distributed.
D. $\mathrm { n } = 18$, the data contain outliers, the variable under consideration is normally distributed.

Accepted Answer

The answer of Provide an appropriate response.
-In which of the...

Question 15

Find the specified t-value.
-For a t-curve with df = 18, find the t-value having area 0.05 to its right.

Accepted Answer

The answer of Find the specified t-value.
-For a t-curve with...

Question 16

Find the confidence interval specified.
-30 students were randomly selected from a large group of students taking a certain calculus test. The mean score for the students in the sample was 78. Assume that $\sigma = 14.5$. Construct a $99 \%$ confidence interval for the mean score, $\mu$, of all students taking the test.

Accepted Answer

The 99% confidence interval is calculated using the formula: $\bar{x} \pm z \times \frac{\sigma}{\sqrt{n}}$. Here, $\bar{x} = 78$, $\sigma = 14.5$, $n = 30$, and $z$ for 99% confidence is approximately 2.576. The interval is $78 \pm 2.576 \times \frac{14.5}{\sqrt{30}}$, which calculates to approximately $71.18$ to $84.82$.

Question 17

Solve the problem.
-A sample of 35 washing machines yields a mean replacement time of 9.0 years. Assuming that $\sigma$ = 2.4 years, find the margin of error in estimating $\mu$ at the 90% level of confidence.

Accepted Answer

The margin of error (E) for estimating the mean ($\mu$) with a certain level of confidence is given by the formula $E = Z \cdot \frac{\sigma}{\sqrt{n}}$, where $Z$ is the Z-score corresponding to the desired level of confidence, $\sigma$ is the population standard deviation, and $n$ is the sample size. For a 90% confidence level, the Z-score is approximately 1.645. Given $\sigma = 2.4$ years and $n = 35$, we calculate $E = 1.645 \cdot \frac{2.4}{\sqrt{35}} \approx 0.7$ years.

Question 18

Find the specified t-value.
-For a t-curve with $d f = 11$, find ${ } { t } _{0.10} { \text {. } }$

Accepted Answer

For a t-distribution with 11 degrees of freedom, the t-value for a one-tailed test with a significance level of 0.10 is approximately 1.363.

Question 19

Find the confidence interval specified. Assume that the population is normally distributed.
-The principal randomly selected six students to take an aptitude test. Their scores were: $\begin{array} { l l l l l l } 
87.4 & 86.9 & 89.9 & 78.3 & 75.1 & 70.6
\end{array}$ Determine a 90% confidence interval for the mean score for all students.

Accepted Answer

The answer of Find the confidence interval specified. Assume that...

Question 20

Provide an appropriate response.
-Give an expression for the confidence level in terms of $\alpha$.

Accepted Answer

The confidence level is typically expressed as $100(1 - \alpha)\%$, where $\alpha$ is the significance level. This formula represents the proportion of times that the confidence interval would contain the true parameter if the experiment were repeated a large number of times.

Quiz 8: Confidence Intervals for One Population Mean

Solve the problem. -A sample of 55 eggs yields a mean weight of 1.76 ounces. Assuming that $\sigma$ = 0.28 ounces, find the margin of error in estimating µ at the 95% level of confidence.

Find the specified t-value. -For a t-curve with $d f = 20$ , find $t _ { 0.01 }$ .

Solve the problem. -Based on a sample of 36 randomly selected years, a 90% confidence interval for the mean annual precipitation in one city is from 46.8 inches to 49.2 inches. Find the margin of error.

Find the specified t-value. -For a t-curve with $\mathrm { df } = 24$ , find $\mathrm { t } _ { 0.005 }$ . $^ { \text {. } }$

Find the specified t-value. -For a t-curve with df = 6, find the two t-values that divide the area under the curve into a middle 0.99 area and two outside areas of

Solve the problem. -A confidence interval for a population mean has a margin of error of 6.4. Determine the length of the confidence interval.

Solve the problem. -A confidence interval for a population mean, µ, has l $\mu _ { ,}$ ength 179.1. Find the margin of error.

Find the specified t-value. -For a t-curve with df = 18, find the t-value having area 0.05 to its right.

Solve the problem. -A sample of 35 washing machines yields a mean replacement time of 9.0 years. Assuming that $\sigma$ = 2.4 years, find the margin of error in estimating $\mu$ at the 90% level of confidence.

Find the specified t-value. -For a t-curve with $d f = 11$ , find ${ } { t } _{0.10} { \text {. } }$

Provide an appropriate response. -Give an expression for the confidence level in terms of $\alpha$ .

Quiz 8: Confidence Intervals for One Population Mean

Solve the problem. -A sample of 55 eggs yields a mean weight of 1.76 ounces. Assuming that σ\sigmaσ = 0.28 ounces, find the margin of error in estimating µ at the 95% level of confidence.

Find the specified t-value. -For a t-curve with df=20d f = 20df=20 , find t0.01t _ { 0.01 }t0.01​ .

Solve the problem. -Based on a sample of 36 randomly selected years, a 90% confidence interval for the mean annual precipitation in one city is from 46.8 inches to 49.2 inches. Find the margin of error.

Find the specified t-value. -For a t-curve with df=24\mathrm { df } = 24df=24 , find t0.005\mathrm { t } _ { 0.005 }t0.005​ . . ^ { \text {. } }.