Question 1

A chi-square distribution is negatively skewed.

Accepted Answer

The chi-square distribution is positively skewed, not negatively skewed.

Question 2

A college believes that 26% of applicants to that school have parents who have remarried. How large a sample is needed to estimate the true proportion of students who
Have parents who have remarried to within 5 percentage points with 95% confidence?

Accepted Answer

To determine the sample size needed for estimating a proportion with a given margin of error and confidence level, we use the formula: n = (Z^2 * p * (1-p)) / E^2. For 95% confidence, Z is 1.96. Here, p = 0.26 and E = 0.05. Plugging these values into the formula gives approximately 296.

Question 3

The following MINITAB output presents a confidence interval for a proportion.
$\begin{array}{crrcc}
\hline \text { Variable } & \mathrm{X} & \mathrm{N} & \text { Sample p } & 95 \% \mathrm{CI} \
\mathrm{x} & 117 & 159 & 0.735849 & (0.667319,0.804379) \
\hline
\end{array}$
Use the information in the display to construct a $98 \%$ confidence interval for $p$.

Accepted Answer

The 95% confidence interval is (0.667319, 0.804379). To construct a 98% confidence interval, we can use the formula:sample proportion +/- 2.326 * sqrt((sample proportion * (1 - sample proportion)) / sample size)Plugging in the values from the output, we get:0.735849 +/- 2.326 * sqrt((0.735849 * (1 - 0.735849)) / 159)= 0.735849 +/- 0.076= (0.659849, 0.811849)Therefore, the 98% confidence interval is (0.659849, 0.811849), which is closest to option A.

Question 4

A recent poll of 700 people who work indoors found that 278 smoke. If the researchers want to be 98% confident of their results to within 3.5 percentage points, how large a
Sample is necessary?

Accepted Answer

To determine the necessary sample size for a given confidence level and margin of error, the formula is used: $n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}$, where $Z$ is the Z-score corresponding to the desired confidence level, $p$ is the proportion of the population that has the characteristic of interest (in this case, the proportion of people who smoke), and $E$ is the margin of error. For a 98% confidence level, the Z-score is approximately 2.33. The proportion $p$ is 278/700, but since we're looking for a general sample size for a population, we use a conservative estimate where $p = 0.5$ for maximum variability. The margin of error $E$ is 0.035 (3.5%). Plugging these values into the formula gives a sample size that rounds to the nearest whole number, which is 1062.

Question 5

The following display from a TI- 84 Plus calculator presents a $99 \%$ confidence interval for proportion.
 a Use the information in the display to construct a 95% confidence interval for p.

Accepted Answer

The answer of The following display from a TI- 84...

Question 6

The following display from a TI-84 Plus calculator presents a $99 \%$ confidence interval for proportion.

a Fill in the blanks: We are ________ confident that the population mean is between _______ and
_______.

Accepted Answer

The answer of The following display from a TI-84 Plus...

Question 7

In a sample of 855 bartenders, 48.0% report hearing complaints from patrons about their jobs. If the margin of error for the proportion of bartenders hearing job complaints is
Percentage points, what is the degree of confidence used?

Accepted Answer

The answer of In a sample of 855 bartenders, 48.0%...

Question 8

John Davis, a manager of a supermarket, wants to estimate the proportion of customers who use food stamps at his store. He has no initial estimate of what the sample
Proportion will be. How large a sample is required to estimate the true proportion to
Within 3 percentage points with 98% confidence?

Accepted Answer

The answer of John Davis, a manager of a supermarket,...

Question 9

A retailer wants to estimate with 99% confidence the number of people who shop at his
store. A previous study showed that 24% of those interviewed had shopped at his store.
He wishes to be accurate within 3% of the true proportion. The minimum sample size
necessary would be 1,100.

Accepted Answer

To estimate the minimum sample size necessary for a given confidence level and margin of error, the formula is $n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}$, where $Z$ is the Z-score corresponding to the desired confidence level (for 99% confidence, $Z \approx 2.576$), $p$ is the estimated proportion (0.24 in this case), and $E$ is the margin of error (0.03). Plugging these values in gives a different number than 1,100, indicating that the statement is false.

Question 10

A report states that 42% of home owners had a vegetable garden. How large a sample is needed to estimate the true proportion of home owners who have vegetable gardens to
Within 4% with 95% confidence?

Accepted Answer

To determine the sample size needed to estimate the true proportion with a certain level of confidence and margin of error, we use the formula for sample size in estimating proportions: $n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}$, where $Z$ is the Z-score corresponding to the confidence level (for 95% confidence, $Z = 1.96$), $p$ is the estimated proportion (0.42 in this case), and $E$ is the margin of error (0.04 for 4%). Plugging in the values: $n = \frac{1.96^2 \cdot 0.42 \cdot (1-0.42)}{0.04^2}$, which calculates to approximately 585. Therefore, a sample size of 585 is needed.

Question 11

A report states that 38% of home owners have a vegetable garden. How large a sample is needed to estimate the true proportion of home owners who have vegetable gardens to
Within 6 percentage points with 98% confidence?

Accepted Answer

To estimate the sample size needed for a proportion with a given confidence level and margin of error, the formula is $n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}$, where $Z$ is the Z-score corresponding to the confidence level, $p$ is the estimated proportion, and $E$ is the margin of error. For a 98% confidence level, the Z-score is approximately 2.33 (from Z-tables). Given $p = 0.38$ and $E = 0.06$, substituting these values gives $n = \frac{(2.33)^2 \cdot 0.38 \cdot (1-0.38)}{(0.06)^2}$, which calculates to approximately 355.8, rounding up to 356.

Question 12

The Academy of Orthopedic Surgeons states that 80% of women wear shoes that are too small for their feet. A researcher wants to be 98% confident that this proportion is within
3 percentage points of the true proportion. How large a sample is necessary?

Accepted Answer

To determine the sample size necessary for a given confidence level and margin of error, the formula for sample size in estimating proportions is used: $n = \frac{{Z^2 \cdot p \cdot (1-p)}}{{E^2}}$, where $Z$ is the Z-score corresponding to the confidence level, $p$ is the estimated proportion, and $E$ is the margin of error. For a 98% confidence level, the Z-score is approximately 2.33 (from Z-tables). Given $p = 0.8$ and $E = 0.03$, substituting these values gives $n = \frac{{(2.33)^2 \cdot 0.8 \cdot (1-0.8)}}{{(0.03)^2}} \approx 966$.

Question 13

The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 80 customers surveyed, 15 ordered only cheese pizza. What is the
99% confidence interval of the true proportion of customers who order only cheese
Pizza?

Accepted Answer

The answer of The Pizza Shop wanted to determine what...

Question 14

$$\text { If } \hat { p } \text { is equal to } 0.88 \text {, then } \hat { q } \text { is equal to }$$________

Accepted Answer

In a probability context, $\hat{p}$ represents the estimated probability of success, and $\hat{q}$ represents the estimated probability of failure. Since probabilities of success and failure must sum to 1, $\hat{q} = 1 - \hat{p}$. Therefore, if $\hat{p} = 0.88$, then $\hat{q} = 1 - 0.88 = 0.12$.

Question 15

A random sample of 75 printers discovered that 25 of them were being used in small businesses . Find the 99% limit for the population proportion of printers that are used in
Small businesses.

Accepted Answer

The 99% confidence interval for a proportion is calculated using the formula $\hat{p} \pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p}$ is the sample proportion, $z$ is the z-score corresponding to the confidence level (for 99% confidence, $z \approx 2.576$), and $n$ is the sample size. Here, $\hat{p} = \frac{25}{75} = \frac{1}{3}$ and $n = 75$. Plugging these values into the formula gives the interval $0.193 < p < 0.473$, which matches choice A.

Question 16

The rounding rule for a confidence interval for a proportion is to round off to three
decimal places.

Accepted Answer

The rounding rule for a confidence interval for a proportion is to round off to three decimal places.

Question 17

The following MINITAB output presents a confidence interval for a proportion.
$\begin{array}{crrcc}
\hline \text { Variable } & \mathrm{X} & \mathrm{N} & \text { Sample p } & 95 \% \mathrm{CI} \
\mathrm{x} & 154 & 191 & 0.806283 & (0.750234,0.862332) \
\hline
\end{array}$
Fill in the blanks: We are_______ confident that the population mean is between_____ _ and

Accepted Answer

The output specifies a 95% confidence interval for the proportion, which ranges from 0.750234 to 0.862332.

Question 18

A researcher wants to construct a 90% confidence interval for the proportion of elementary school students in Seward County who receive free or reduced-price school
Lunches. A state-wide survey indicates that the proportion is 0.50. Using this estimate,
What sample size is needed so that the confidence interval will have a margin of error of
0)1?

Accepted Answer

The sample size needed for a proportion can be calculated using the formula $n = \left(\frac{Z^2 \cdot p \cdot (1-p)}{E^2}\right)$, where $Z$ is the Z-score corresponding to the desired confidence level (for 90% confidence, $Z = 1.645$), $p$ is the estimated proportion (0.50 in this case), and $E$ is the desired margin of error (0.1). Plugging in the values: $n = \left(\frac{1.645^2 \cdot 0.50 \cdot (1-0.50)}{0.1^2}\right) = 67.6025$, which rounds up to 68 to ensure the sample is large enough.

Question 19

A researcher wants to construct a 99% confidence interval for the proportion of elementary school students in Seward County who receive free or reduced-price school
Lunches. What sample size is needed so that the confidence interval will have a margin of
Error of 0.07?

Accepted Answer

To calculate the sample size for a proportion with a given margin of error, we use the formula: n = (Z^2 * p * (1-p)) / E^2. For a 99% confidence interval, Z is approximately 2.576. Assuming p = 0.5 for maximum variability, and E = 0.07, the calculation gives a sample size of approximately 339.

Question 20

The symbol for the sample proportion is .

Accepted Answer

The answer of The symbol for the sample proportion is...

Quiz 7: Confidence Intervals and Sample Size

A chi-square distribution is negatively skewed.

A college believes that 26% of applicants to that school have parents who have remarried. How large a sample is needed to estimate the true proportion of students who Have parents who have remarried to within 5 percentage points with 95% confidence?

A recent poll of 700 people who work indoors found that 278 smoke. If the researchers want to be 98% confident of their results to within 3.5 percentage points, how large a Sample is necessary?

In a sample of 855 bartenders, 48.0% report hearing complaints from patrons about their jobs. If the margin of error for the proportion of bartenders hearing job complaints is Percentage points, what is the degree of confidence used?

John Davis, a manager of a supermarket, wants to estimate the proportion of customers who use food stamps at his store. He has no initial estimate of what the sample Proportion will be. How large a sample is required to estimate the true proportion to Within 3 percentage points with 98% confidence?

A retailer wants to estimate with 99% confidence the number of people who shop at his store. A previous study showed that 24% of those interviewed had shopped at his store. He wishes to be accurate within 3% of the true proportion. The minimum sample size necessary would be 1,100.

A report states that 42% of home owners had a vegetable garden. How large a sample is needed to estimate the true proportion of home owners who have vegetable gardens to Within 4% with 95% confidence?

A report states that 38% of home owners have a vegetable garden. How large a sample is needed to estimate the true proportion of home owners who have vegetable gardens to Within 6 percentage points with 98% confidence?

The Academy of Orthopedic Surgeons states that 80% of women wear shoes that are too small for their feet. A researcher wants to be 98% confident that this proportion is within 3 percentage points of the true proportion. How large a sample is necessary?

The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 80 customers surveyed, 15 ordered only cheese pizza. What is the 99% confidence interval of the true proportion of customers who order only cheese Pizza?

$\text { If } \hat { p } \text { is equal to } 0.88 \text {, then } \hat { q } \text { is equal to }$ ________

A random sample of 75 printers discovered that 25 of them were being used in small businesses . Find the 99% limit for the population proportion of printers that are used in Small businesses.

The rounding rule for a confidence interval for a proportion is to round off to three decimal places.

A researcher wants to construct a 99% confidence interval for the proportion of elementary school students in Seward County who receive free or reduced-price school Lunches. What sample size is needed so that the confidence interval will have a margin of Error of 0.07?

The symbol for the sample proportion is .

Quiz 7: Confidence Intervals and Sample Size

A chi-square distribution is negatively skewed.

A college believes that 26% of applicants to that school have parents who have remarried. How large a sample is needed to estimate the true proportion of students who Have parents who have remarried to within 5 percentage points with 95% confidence?

A recent poll of 700 people who work indoors found that 278 smoke. If the researchers want to be 98% confident of their results to within 3.5 percentage points, how large a Sample is necessary?

The following display from a TI- 84 Plus calculator presents a 99%99 \%99% confidence interval for proportion. a Use the information in the display to construct a 95% confidence interval for p.

The following display from a TI-84 Plus calculator presents a 99%99 \%99% confidence interval for proportion. a Fill in the blanks: We are ________ confident that the population mean is between _______ and _______.

In a sample of 855 bartenders, 48.0% report hearing complaints from patrons about their jobs. If the margin of error for the proportion of bartenders hearing job complaints is Percentage points, what is the degree of confidence used?

John Davis, a manager of a supermarket, wants to estimate the proportion of customers who use food stamps at his store. He has no initial estimate of what the sample Proportion will be. How large a sample is required to estimate the true proportion to Within 3 percentage points with 98% confidence?

A retailer wants to estimate with 99% confidence the number of people who shop at his store. A previous study showed that 24% of those interviewed had shopped at his store. He wishes to be accurate within 3% of the true proportion. The minimum sample size necessary would be 1,100.

A report states that 42% of home owners had a vegetable garden. How large a sample is needed to estimate the true proportion of home owners who have vegetable gardens to Within 4% with 95% confidence?

A report states that 38% of home owners have a vegetable garden. How large a sample is needed to estimate the true proportion of home owners who have vegetable gardens to Within 6 percentage points with 98% confidence?

The Academy of Orthopedic Surgeons states that 80% of women wear shoes that are too small for their feet. A researcher wants to be 98% confident that this proportion is within 3 percentage points of the true proportion. How large a sample is necessary?

The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 80 customers surveyed, 15 ordered only cheese pizza. What is the 99% confidence interval of the true proportion of customers who order only cheese Pizza?

If p^ is equal to 0.88, then q^ is equal to \text { If } \hat { p } \text { is equal to } 0.88 \text {, then } \hat { q } \text { is equal to } If p^​ is equal to 0.88, then q^​ is equal to ________

A random sample of 75 printers discovered that 25 of them were being used in small businesses . Find the 99% limit for the population proportion of printers that are used in Small businesses.

The rounding rule for a confidence interval for a proportion is to round off to three decimal places.

A researcher wants to construct a 99% confidence interval for the proportion of elementary school students in Seward County who receive free or reduced-price school Lunches. What sample size is needed so that the confidence interval will have a margin of Error of 0.07?

The symbol for the sample proportion is .

$\text { If } \hat { p } \text { is equal to } 0.88 \text {, then } \hat { q } \text { is equal to }$ ________