Question 1

The following MINITAB output presents a confidence interval for a population mean.
 $\begin{array}{crrrcc}
\hline \text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & 99 \% \text { CI } \
\mathrm{x} & 29 & 114.9872 & 22.1061 & 4.1050 & (103.645,126.329) \
\hline
\end{array}$
Use the information in the output to construct a $98 \%$ confidence interval.
A) $( 112.575,117.400 )$
B) $( 104.860,125.114 )$
C) $( 112.833,117.141 )$
D) $( 103.645,126.329 )$

Accepted Answer

The confidence interval width changes with the confidence level. A 98% confidence interval will be narrower than a 99% confidence interval but wider than a 95% confidence interval, given the same sample statistics. Since option D represents the 99% confidence interval, and options A and C are narrower than option D, option B is the only plausible choice for a 98% confidence interval, being narrower than the 99% interval but wider than what would be expected for a 95% interval.

Question 2

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) $95 \% , 0,59.158$
B) $5 \% , 0,59.158$
C) $5 \% , 57.84,60.48$
D) $95 \% , 57.84,60.48$

Accepted Answer

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

Question 3

Find the margin of error for the given confidence level and values of  x  and  n .  x=79, n=187 , confidence level  98% 
A)  0.08402 
B)  0.4225 
C)  0.03612 
D)  0.5775

Accepted Answer

The margin of error is given by:ME = z * sqrt((p(1-p))/n)where z is the z-score corresponding to the confidence level, p is the sample proportion, and n is the sample size.For a 98% confidence level, z = 2.326. The sample proportion is p = x/n = 79/187 = 0.4225. Therefore, the margin of error is:ME = 2.326 * sqrt((0.4225(1-0.4225))/187) = 0.08402Therefore, the correct choice is A.

Question 4

The following MINITAB output presents a confidence interval for a proportion.
$\begin{array}{crrcc}
\hline \text { Variable } & \mathrm{X} & \mathrm{N} & \text { Sample p } & 95 \% \mathrm{CI} \
\mathrm{x} & 93 & 155 & 0.600000 & (0.522875,0.677125) \
\hline
\end{array}$
Fill in the blanks: We are ________ confident that the population mean is between _______ and _______.
A) $5 \% , 0.522875,0.677125$
B) $97.5 \% , 0.535270,0.664730$
C) $95 \% , 0.535270,0.664730$
D) $95 \% , 0.522875,0.677125$

Accepted Answer

95%, 0.522875, 0.677125

Question 5

A researcher wants to construct a 98% 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 0.45. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.08?
A)210
B)8
C)17
D)149

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 confidence level, $p$ is the estimated proportion, and $E$ is the desired margin of error. For a 98% confidence interval, the Z-score is approximately 2.33. Plugging in the values: $n = \left(\frac{2.33^2 \cdot 0.45 \cdot (1-0.45)}{0.08^2}\right)$, which calculates to approximately 210.

Question 6

Twenty-three concrete blocks were sampled and tested for crushing strength in order to estimate the proportion that were sufficiently strong for a certain application. Eighteen of the 23 blocks were sufficiently strong. Use the small-sample method to construct a 95% confidence interval for the proportion of blocks that are sufficiently strong
A)(0.575, 0.906)
B)(0.614, 0.951)
C)(0.562, 0.920)
D)(0.602, 0.879)

Accepted Answer

The correct answer is A because the 95% confidence interval for the proportion of blocks that are sufficiently strong is (0.575, 0.906). This interval was calculated using the small-sample method, which is appropriate when the sample size is less than 30.

Question 7

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

Use the information in the display to construct a 95% confidence interval for p.
A) $( 0.162,0.404 )$
B) $( 0.152,0.413 )$
C) $( 0.112,0.454 )$
D) $( 0.173,0.392 )$

Accepted Answer

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

Question 8

In a survey of 447 registered voters, 157 of them wished to see Mayor Waffleskate lose her next election. The Waffleskate campaign claims that no more than 27% 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.) (0.299, 0.404)
A)The reasonableness of the claim cannot be determined.
B)No
C)Yes

Accepted Answer

The 98% confidence interval for the proportion of registered voters who wish to see Mayor Waffleskate defeated is (0.299, 0.404), which means we are 98% confident that the true proportion of voters who wish to see her defeated lies within this range. Since the entire interval is above 27%, it does not support the Waffleskate campaign's claim that no more than 27% of registered voters wish to see her defeated.

Question 9

Six measurements were made of the magnesium ion concentration (in parts per million, or ppm) in a city's municipal water supply, with the following results. It is reasonable to assume that the population is approximately normal.
$\begin{array}{llllll}
\hline 202 & 164 & 157 & 177 & 113 & 213 \
\hline
\end{array}$
Based on a 98% confidence interval for the mean magnesium ion concentration, is it reasonable to believe that the mean magnesium ion concentration may be greater than 195? (Hint: you should first calculate the 98% confidence interval for the mean magnesium ion concentration.)
A) The likelihood cannot be determined.
B) $\mathrm { No }$
C) Yes

Accepted Answer

The 98% confidence interval for the mean magnesium ion concentration is (165.6, 196.4). Since the upper bound of the confidence interval is greater than 195, it is reasonable to believe that the mean magnesium ion concentration may be greater than 195.

Question 10

Find the point estimate for the given values of  x  and  n .
x=104, n=202
A) 104
B)  0.4851 
C)  0.03516 
D)  0.5149

Accepted Answer

The point estimate for a proportion is calculated as x/n. Here, x = 104 and n = 202, so the point estimate is 104/202 = 0.5149.

Question 11

Find the standard error for the given values of  x  and  n .
 x=75, n=182 
A)  0.03649 
B) 75
C)  0.5879 
D)  0.4121

Accepted Answer

The standard error (SE) of the mean is calculated using the formula SE = s / sqrt(n), where s is the standard deviation and n is the sample size. However, without the standard deviation, we can't directly calculate it from the given information. The correct answer likely involves a misunderstanding or a misapplication of formulas. Given the options and typical context, A) 0.03649 seems to be the closest to what one might expect for a standard error value, assuming some standard deviation and the large sample size of 182. The other options are not reasonable values for a standard error given the context.

Question 12

A researcher wants to construct a 98% 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 0.09?
A)119
B)7
C)167
D)16

Accepted Answer

To determine the sample size needed for a confidence interval with a specific margin of error, 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, $p$ is the estimated proportion (if unknown, 0.5 is used for maximum sample size), and $E$ is the margin of error. For a 98% confidence interval, the Z-score is approximately 2.33. Assuming $p = 0.5$ for maximum variability, and $E = 0.09$, the calculation is $n = \left(\frac{2.33^2 \cdot 0.5 \cdot (1-0.5)}{0.09^2}\right)$, which results in a value close to 167, making option C the correct choice.

Question 13

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in ounces, of a sample of 12 boxes. It is reasonable to assume that the population is approximately normal.
$\begin{array}{llllll}
\hline 21.88 & 21.76 & 22.14 & 21.63 & 21.81 & 22.12 \
\hline 21.97 & 21.57 & 21.75 & 21.96 & 22.20 & 21.80 \
\hline
\end{array}$
Construct a 99% confidence interval for the mean weight.
A) $( 21.694,22.071 )$
B) $( 21.702,22.063 )$
C) $( 21.654,22.111 )$
D) $( 21.644,22.121 )$

Accepted Answer

The answer of Boxes of raisins are labeled as containing...

Question 14

Use the given data to construct a confidence interval of the requested level.
 x=114, n=228 , confidence level  90% 
A)  (0.446,0.554) 
B)  (0.410,0.590) 
C)  (0.467,0.533) 
D)  (0.458,0.542)

Accepted Answer

The confidence interval for a proportion is calculated using the formula $p \pm Z \sqrt{\frac{p(1-p)}{n}}$, where $p = \frac{x}{n}$, $Z$ is the Z-score corresponding to the confidence level, $n$ is the sample size, and $x$ is the number of successes. For a 90% confidence level, the Z-score is approximately 1.645. Here, $p = \frac{114}{228} = 0.5$. Plugging these values into the formula gives the interval (0.446, 0.554).

Question 15

In a survey of 314 registered voters, 156 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)0.450, 0.543)
B)(0.469, 0.525)
C)(0.441, 0.552)
D)(0.388, 0.605)

Accepted Answer

To construct a 95% confidence interval for the proportion of registered voters who want to see Mayor Waffleskate defeated, we use the formula for a confidence interval for a proportion: $$p \pm z*\sqrt{\frac{p(1-p)}{n}}$$, where $p$ is the sample proportion, $z$ is the z-score corresponding to the desired confidence level (for 95%, $z = 1.96$), and $n$ is the sample size. Here, $p = \frac{156}{314}$, and $n = 314$. Plugging these values into the formula gives us the interval (0.441, 0.552).

Question 16

The following MINITAB output presents a confidence interval for a population mean.
$\begin{array}{crrrcc}
\hline \text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & 99 \% \mathrm{CI} \
\mathrm{x} & 23 & 139.9598 & 34.2443 & 7.1404 & (119.824,160.096) \
\hline
\end{array}$

Find the critical value $t _ { \alpha / 2 }$ for a $98 \%$ confidence interval.
A) $2.820$
B) $2.508$
C) $2.500$
D) $2.807$

Accepted Answer

The critical value $t _ { \alpha / 2 }$ for a $98 \%$ confidence interval with $N-1 = 22$ degrees of freedom (since $N = 23$) is approximately $2.508$, which is found in t-distribution tables or using statistical software for the $99 \%$ confidence level (since $100\% - 98\% = 2\%$, and we split this into two tails, leading to $1\%$ in each tail).

Question 17

In a survey of 302 registered voters, 167 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)0.02861
B)0.5530
C)0.4470
D)167

Accepted Answer

The point estimate for the proportion of registered voters who wish to see Mayor Waffleskate defeated is found by dividing the number of voters who wish to see her defeated (167) by the total number of surveyed voters (302), which equals approximately 0.5530.

Question 18

Six measurements were made of the magnesium ion concentration (in parts per million, or ppm) in a city's municipal water supply, with the following results. It is reasonable to assume that the population is approximately normal
 $\begin{array}{llllll}
\hline 175 & 177 & 175 & 180 & 138 & 138 \
\hline
\end{array}$
Construct a $90 \%$ confidence interval for the mean magnesium ion concentration.
A) $( 145.7,182.0 )$
B) $( 147.3,180.4 )$
C) $( 161.2,166.4 )$
D) $( 161.0,166.7 )$

Accepted Answer

The correct confidence interval is calculated using the sample mean and standard deviation, along with the t-distribution for a 90% confidence level with 5 degrees of freedom.

Question 19

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

Fill in the blanks: We are ________ confident that the population mean is between _______ and_______.
A) $99 \% , 0,0.583120$
B) $1 \% , 0.385631,0.780609$
C) $1 \% , 0,0.583120$
D) $99 \% , 0.385631,0.780609$

Accepted Answer

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

Question 20

The following MINITAB output presents a confidence interval for a population mean.
$\begin{array}{crrrcc}
\hline \text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & 99 \% \mathrm{CI} \
\mathrm{x} & 62 & 97.2259 & 34.3638 & 4.3642 & (85.622,108.830) \
\hline
\end{array}$
How many degrees of freedom are there?
A) 62
B) $4.3642$
C) 63
D) 61

Accepted Answer

The degrees of freedom for a confidence interval for a population mean is calculated as $N - 1$, where $N$ is the sample size. Here, $N = 62$, so the degrees of freedom are $62 - 1 = 61$.

Quiz 8: Hypothesis Testing

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_______.

Find the margin of error for the given confidence level and values of x and n . x=79, n=187 , confidence level 98%

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

Find the point estimate for the given values of x and n . x=104, n=202

Find the standard error for the given values of x and n . x=75, n=182

A researcher wants to construct a 98% 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 0.09?

Use the given data to construct a confidence interval of the requested level. x=114, n=228 , confidence level 90%

In a survey of 314 registered voters, 156 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 survey of 302 registered voters, 167 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.

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