Question 1

7 squirrels were found to have an average weight of 8.7 ounces with a sample standard deviation is 1.1. Find the 95% confidence interval of the true mean weight. 
A) $6.0 < \mu < 11.4$
B) $7.7 < \mu < 9.7$
C) $8.3 < \mu < 9.1$
D) $7.9 < \mu < 9.5$

Accepted Answer

To find the 95% confidence interval for the true mean weight of the squirrels, we use the formula for the confidence interval of the mean with a t-distribution (since the sample size is small and the population standard deviation is unknown): $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $t_{\alpha/2}$ is the t-score for 95% confidence and $n-1$ degrees of freedom, $s$ is the sample standard deviation, and $n$ is the sample size.Given: $\bar{x} = 8.7$, $s = 1.1$, and $n = 7$.Degrees of freedom $df = n - 1 = 6$.For a 95% confidence interval and 6 degrees of freedom, the t-score (from a t-table or calculator) is approximately 2.447.Plugging these values into the formula gives us the margin of error: $E = 2.447 \cdot \frac{1.1}{\sqrt{7}} \approx 1.0$.Thus, the confidence interval is $8.7 \pm 1.0$, which is $7.7 < \mu < 9.7$.

Question 2

A survey of 800 women shoppers found that 17% of them shop on impulse. What is the 98% confidence interval for the true proportion of women shoppers who shop on
Impulse? 
A) $0.139 < p < 0.201$
B) $0.136 < p < 0.204$
C) $0.167 < p < 0.173$
D) $0.144 < p < 0.196$

Accepted Answer

To calculate the 98% confidence interval for the proportion of women shoppers who shop on impulse, we use the formula for the confidence interval of a proportion: $\hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p}$ is the sample proportion, $Z_{\alpha/2}$ is the Z-score corresponding to the confidence level, and $n$ is the sample size. Given $\hat{p} = 0.17$ (17%) and $n = 800$, and knowing that the Z-score for a 98% confidence level is approximately 2.33, we calculate the margin of error and add/subtract it from the sample proportion to find the confidence interval. The calculation yields the interval $0.139 < p < 0.201$, which matches choice A.

Question 3

Find the standard error for the given values of $x$ and $n$.
$$x = 125 , n = 259$$
A) 125
B) 0.483
C) 0.031
D) 0.517

Accepted Answer

The standard error (SE) for a proportion is calculated using the formula $ SE = \sqrt{\frac{p(1-p)}{n}} $, where $ p = \frac{x}{n} $. Here, $ p = \frac{125}{259} \approx 0.4827 $. Plugging in the values, $ SE \approx \sqrt{\frac{0.4827 \times (1-0.4827)}{259}} \approx 0.031 $.

Question 4

The following MINITAB output presents a confidence interval for a population mean.
$\begin{array}{crrrrc}
\hline \text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & 99 \% \mathrm{CI} \
\mathrm{x} & 72 & 84.4238 & 33.8186 & 3.9856 & (73.874,94.973) \
\hline
\end{array}$
How many degrees of freedom are there?
A) 3.9856
B) 72
C) 73
D) 71

Accepted Answer

The degrees of freedom for a confidence interval for a population mean, when the sample standard deviation is used as an estimate for the population standard deviation, is calculated as the sample size minus one. Here, N = 72, so the degrees of freedom are 72 - 1 = 71.

Question 5

The following MINITAB output presents a confidence interval for a population mean.
$\begin{array}{crrrcc}
\hline \text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & 99 \% \text { CI } \
x & 30 & 89.4928 & 33.3397 & 6.0870 & (72.693,106.293) \
\hline
\end{array}$
Find the critical value $t _ { \alpha / 2 }$ for a $98 \%$ confidence interval.
A) 2.462
B) 2.750
C) 2.760
D) 2.457

Accepted Answer

The critical value $t _ { \alpha / 2 }$ for a 98% confidence interval with 29 degrees of freedom (since $N = 30$) is closest to 2.462, which can be found in t-distribution tables or using statistical software for the 99% confidence level (since the question mistakenly asks for 98% but provides data for 99%).

Question 6

The following MINITAB output presents a confidence interval for a population mean.
$\begin{array}{crrccc}
\hline \text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & 99 \% \mathrm{CI} \
\mathrm{x} & 28 & 108.1268 & 30.4888 & 5.7618 & (92.161,124.093) \
\hline
\end{array}$
Use the information in the output to construct a $98 \%$ confidence interval.
A) (92.161, 124.093)
B) (105.235, 111.018)
C) (93.878, 122.376)
D) (105.546, 110.707)

Accepted Answer

The 99% confidence interval is wider than the 98% confidence interval due to the higher confidence level requiring a larger margin of error. Since option C provides a range that is narrower than the 99% confidence interval but still reasonably wide, it is the correct choice. Options B and D are too narrow, and option A is the original 99% confidence interval.

Question 7

In a survey of 246 registered voters, 132 of them wished to see Mayor Waffleskate lose her next election. Find a point estimate for the proportion of registered voters who wish
To see Mayor Waffleskate defeated.
A) 0.463
B) 132
C) 0.032
D) 0.537

Accepted Answer

The point estimate for the proportion is calculated by dividing the number of voters who wish to see the mayor defeated (132) by the total number of surveyed voters (246), which equals 0.537.

Question 8

Find the margin of error for the given confidence level and values of $x$ and $n$. $x = 111 , n = 194$, confidence level $99 \%$
A) 0.428
B) 0.572
C) r0.036
D) 0.093

Accepted Answer

The margin of error is calculated using the formula for a proportion: $ ME = Z \times \sqrt{\frac{p(1-p)}{n}} $, where $ p = \frac{x}{n} $ and $ Z $ is the Z-score for the confidence level. For 99% confidence, $ Z \approx 2.576 $. Here, $ p = \frac{111}{194} \approx 0.572 $. Plugging in the values, the margin of error is approximately 0.093.

Question 9

A sample of 400 racing cars showed that 80 of them cost over $700,000. What is
the 99% confidence interval for the true proportion of racing cars that cost over
$700,000?

Accepted Answer

The answer of A sample of 400 racing cars showed...

Question 10

In a survey of 305 registered voters, 130 of them wished to see Mayor Waffleskate lose her next election. Construct a 95% confidence interval for the proportion of registered
Voter who want to see Mayor Waffleskate defeated. 
A) $0.371 < p < 0.482$
B) $0.380 < p < 0.473$
C) $0.317 < p < 0.535$
D) $0.398 < p < 0.455$

Accepted Answer

The sample size is 305 and 130 of them wished to see Mayor Waffleskate lose her next election. So the sample proportion is 130/305 = 0.426. The standard error of the proportion is sqrt(0.426 * 0.574 / 305) = 0.031. The 95% confidence interval is 0.426 +/- 1.96 * 0.031 = (0.371, 0.482).

Question 11

In a sample of 45 mice, a biologist found that 40% were able to run a maze in 30 seconds or less. Find the 95% limit for the population proportion of mice who can run a maze in
30 seconds or less. 
A) $25.7 \% < p < 54.3 \%$
B) $28.0 \% < p < 52.0 \%$
C) $18.5 \% < p < 61.5 \%$
D) $32.8 \% < p < 47.2 \%$

Accepted Answer

To find the 95% confidence interval for the population proportion, we use the formula for the confidence interval of a proportion: $$p \pm Z \sqrt{\frac{p(1-p)}{n}}$$, where $p$ is the sample proportion, $Z$ is the Z-score for the desired confidence level (1.96 for 95%), and $n$ is the sample size. Given that 40% (or 0.4) of the 45 mice could run the maze in 30 seconds or less, we calculate the confidence interval as follows:$$0.4 \pm 1.96 \sqrt{\frac{0.4(0.6)}{45}}$$Calculating the margin of error:$$1.96 \sqrt{\frac{0.4(0.6)}{45}} \approx 0.147$$Thus, the 95% confidence interval is:$$0.4 - 0.147 < p < 0.4 + 0.147$$$$0.253 < p < 0.547$$Converting to percentages:$$25.3\% < p < 54.7\%$$The closest option to our calculated interval is:$25.7 \% < p < 54.3 \%$.

Question 12

A recent study of 750 internet users in Europe found that 35% of internet users were women. What is the 95% confidence interval of the true proportion of women in Europe
Who use the internet? 
A) $0.309 < p < 0.391$
B) $0.321 < p < 0.379$
C) $0.316 < p < 0.384$
D) $0.349 < p < 0.351$

Accepted Answer

The confidence interval is calculated using the formula for a proportion: $\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p} = 0.35$, $Z = 1.96$ for 95% confidence, and $n = 750$. This calculation results in the interval $0.316 < p < 0.384$.

Question 13

In a study of 100 new cars, 29 are white. Find $\hat { p }$ and $\hat { q }$, where $\hat { p }$ is the proportion of new cars that are white.
A) $\hat { p } = 0.29 , \hat { q } = 0.29$
B) $\hat { p } = 0.71 , \hat { q } = 0.29$
C) $\hat { p } = 0.29 , \hat { q } = 0.71$
D) $\hat { p } = 0.71 , \hat { q } = 0.71$

Accepted Answer

$\hat { p }$ is the proportion of new cars that are white, which is 29 out of 100, or 0.29. $\hat { q }$ is the proportion of new cars that are not white, which is 1 - $\hat { p }$ = 0.71.

Question 14

A random sample of 50 voters found that 46% were going to vote for a certain candidate. Find the 95% limit for the population proportion of voters who will vote for that
Candidate. 
A) $34.4 \% < p < 57.6 \%$
B) $32.2 \% < p < 59.8 \%$
C) $39.1 \% < p < 52.9 \%$
D) $33.3 \% < p < 58.7 \%$

Accepted Answer

The answer of A random sample of 50 voters found...

Question 15

Find the point estimate for the given values of $x$ and $n$.
$$x = 122 , n = 201$$
A) 0.607
B) 0.034
C) 122
D) 0.393

Accepted Answer

The point estimate for a proportion is calculated as $p = \frac{x}{n}$, where $x$ is the number of successes and $n$ is the total number of trials. Thus, for $x = 122$ and $n = 201$, the point estimate is $p = \frac{122}{201} \approx 0.607$.

Question 16

The following display from a TI-84 Plus calculator presents a $95 \%$ confidence interval.

Fill in the blanks: We are ________ confident that the population mean is between _______ and _______.
A) 5%, 57.84, 60.48
B) 95%, 0, 59.158
C) 5%, 0, 59.158
D) 95%, 57.84, 60.48

Accepted Answer

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

Question 17

Use the given data to construct a confidence interval of the requested level. $x = 98 , n = 223$, confidence level 99\%
A) $0.219 < p < 0.660$
B) $0.354 < p < 0.525$
C) $0.362 < p < 0.517$
D) $0.406 < p < 0.473$

Accepted Answer

The answer of Use the given data to construct a...

Question 18

It was found that in a sample of 90 teenage boys, 70% of them have received speeding tickets. What is the 90% confidence interval of the true proportion of teenage boys who
Have received speeding tickets? 
A) $0.584 < p < 0.830$
B) $0.591 < p < 0.812$
C) $0.620 < p < 0.780$
D) $0.615 < p < 0.805$

Accepted Answer

The answer of It was found that in a sample...

Question 19

In a survey of 445 registered voters, 131 of them wished to see Mayor Waffleskate lose her next election. The Waffleskate campaign claims that no more than 30% of registered
Voters wish to see her defeated. Does the 98% confidence interval for the proportion
Support this claim? (Hint: you should first construct the 98% confidence interval for the
Proportion of registered voters who whish to see Waffleskate defeated.)
(0)244, 0.345)
A) Yes
B) No
C) The reasonableness of the claim cannot be determined.

Accepted Answer

The 98% confidence interval for the proportion of registered voters who wish to see Waffleskate defeated is (0.244, 0.345). Since the upper bound of the confidence interval (0.345) is greater than 0.30, we can conclude that the Waffleskate campaign's claim is not supported by the data.

Question 20

The winning team's score in 13 high school basketball games was recorded. If the sample mean is 54.3 points and the sample standard deviation is 13.0 points, find the
98% confidence interval of the true mean. 
A) $50.7 < \mu < 57.9$
B) $45.9 < \mu < 62.7$
C) $44.6 < \mu < 64.0$
D) $35.0 < \mu < 73.6$

Accepted Answer

The 98% confidence interval is calculated using the formula: mean ± (critical value) * (standard deviation/√n). For 98% confidence with 12 degrees of freedom, the t-value is approximately 2.681. Thus, the interval is 54.3 ± 2.681 * (13/√13), which results in approximately 44.6 to 64.0.

Quiz 7: Confidence Intervals and Sample Size

7 squirrels were found to have an average weight of 8.7 ounces with a sample standard deviation is 1.1. Find the 95% confidence interval of the true mean weight.

A survey of 800 women shoppers found that 17% of them shop on impulse. What is the 98% confidence interval for the true proportion of women shoppers who shop on Impulse?

Find the standard error for the given values of $x$ and $n$ . $x = 125 , n = 259$

In a survey of 246 registered voters, 132 of them wished to see Mayor Waffleskate lose her next election. Find a point estimate for the proportion of registered voters who wish To see Mayor Waffleskate defeated.

Find the margin of error for the given confidence level and values of $x$ and $n$ . $x = 111 , n = 194$ , confidence level $99 \%$

A sample of 400 racing cars showed that 80 of them cost over $700,000. What is the 99% confidence interval for the true proportion of racing cars that cost over $700,000?

In a survey of 305 registered voters, 130 of them wished to see Mayor Waffleskate lose her next election. Construct a 95% confidence interval for the proportion of registered Voter who want to see Mayor Waffleskate defeated.

In a sample of 45 mice, a biologist found that 40% were able to run a maze in 30 seconds or less. Find the 95% limit for the population proportion of mice who can run a maze in 30 seconds or less.

A recent study of 750 internet users in Europe found that 35% of internet users were women. What is the 95% confidence interval of the true proportion of women in Europe Who use the internet?

In a study of 100 new cars, 29 are white. Find $\hat { p }$ and $\hat { q }$ , where $\hat { p }$ is the proportion of new cars that are white.

A random sample of 50 voters found that 46% were going to vote for a certain candidate. Find the 95% limit for the population proportion of voters who will vote for that Candidate.

Find the point estimate for the given values of $x$ and $n$ . $x = 122 , n = 201$

Use the given data to construct a confidence interval of the requested level. $x = 98 , n = 223$ , confidence level 99\%

It was found that in a sample of 90 teenage boys, 70% of them have received speeding tickets. What is the 90% confidence interval of the true proportion of teenage boys who Have received speeding tickets?

The winning team's score in 13 high school basketball games was recorded. If the sample mean is 54.3 points and the sample standard deviation is 13.0 points, find the 98% confidence interval of the true mean.