Question 1

Assume that you plan to use a significance level of $\alpha = 0.05$ to test the claim that $p _ { 1 } = p _ { 2 }$ Use the given sample sizes and numbers of successes to find the  P -value for the hypothesis test.
$\begin{array} { l l } 
n _ { 1 } = 100 & n _ { 2 } = 100 \
x _ { 1 } = 38 & x _ { 2 } = 40
\end{array}$
A)  0.1610 
B)  0.2130 
C)  0.0412 
D)  0.7718

Accepted Answer

The answer of Assume that you plan to use a...

Question 2

In the context of a hypothesis test for two proportions, which of the following statements about the pooled sample proportion, $\bar { p }$ is/are true?
I. It estimates the common value of $p _ { 1 } \text { and } p _ { 2 }$ under the assumption of equal proportions.
II. It is obtained by averaging the two sample proportions $\hat { p } _ { 1 } \text { and } \hat { p } _ { 2 }$
III. It is equal to the proportion of successes in both samples combined.
A) I and II
B) III only
C) I, II, and III
D) I and III

Accepted Answer

The pooled sample proportion, $\bar { p }$, is used under the assumption that the two populations have the same proportion (statement I is true). It is calculated by combining the successes and the total number of trials from both samples, effectively finding the proportion of successes in both samples combined (statement III is true). It is not simply an average of the two sample proportions (statement II is false), which makes option D correct.

Question 3

Assume that you want to test the claim that the paired sample data come from a population for which the mean difference is $\mu _ { d } = 0 .$
Compute the value of the  t  test statistics. Round intermediate calculations to four decimal places as needed and final answers to three decimal places as needed.
$\begin{array} { l | l l l l l } 
x & 9 & 6 & 7 & 5 & 12 \
\hline y & 6 & 8 & 3 & 6 & 7
\end{array}$
A)  t=0.578 
B)  t=2.890 
C)  t=1.292 
D)  t=0.415

Accepted Answer

To calculate the t-test statistic for paired sample data, we first find the differences between each pair of x and y, calculate the mean of these differences ($\bar{d}$), and the standard deviation of these differences (s_d). Then, we use the formula for the t-test statistic for paired samples: $t = \frac{\bar{d} - \mu_d}{s_d / \sqrt{n}}$, where $\mu_d = 0$ (the hypothesized mean difference), and $n$ is the number of pairs.Differences (d) = x - y = [3, -2, 4, -1, 5]Mean of differences ($\bar{d}$) = (3 - 2 + 4 - 1 + 5) / 5 = 9 / 5 = 1.8Variance of differences = [((3-1.8)^2 + (-2-1.8)^2 + (4-1.8)^2 + (-1-1.8)^2 + (5-1.8)^2) / (5-1)] = [(1.44 + 14.44 + 4.84 + 7.84 + 10.24) / 4] = 38.8 / 4 = 9.7Standard deviation of differences (s_d) = sqrt(9.7) ≈ 3.1145t = (1.8 - 0) / (3.1145 / sqrt(5)) ≈ 1.8 / (3.1145 / 2.236) ≈ 1.8 / 1.392 ≈ 1.292Therefore, the correct choice is C) t=1.292.

Question 4

Find the number of successes x suggested by the given statement. A computer manufacturer_ randomly selects 2680 of its computers for quality assurance and finds that 1.98% of these
Computers are found to be defective.
A)56
B)51
C)58
D)53

Accepted Answer

The answer of Find the number of successes x suggested...

Question 5

A test of abstract reasoning is given to a random sample of students before and after they_ completed a formal logic course. The results are given below. Construct a 95% confidence
Interval for the mean difference between the before and after scores. $\begin{array} { l l l l l l l l l l l l } 
\text { Before } & 74 & 83 & 75 & 88 & 84 & 63 & 93 & 84 & 91 & 77 \
\hline \text { After } & 73 & 77 & 70 & 77 & 74 & 67 & 95 & 83 & 84 & 75
\end{array}$

A) $1.0 < \mu _ { d } < 6.4$
B) $0.2 < \mu _ { d } < 7.2$
C) $0.8 < \mu _ { d } < 6.6$
D) $1.2 < \mu _ { d } < 8.7$

Accepted Answer

The answer of A test of abstract reasoning is given...

Question 6

Determine whether the samples are dependent or independent. The effectiveness of a_ headache medicine is tested by measuring the intensity of a headache in patients before and
After drug treatment. The data consist of before and after intensities for each patient.
A)Dependent samples
B)Independent samples

Accepted Answer

The samples are dependent because the same subjects are measured twice, before and after the treatment, making their before and after measurements related to each other.

Question 7

When performing a hypothesis test for the ratio of two population variances, the upper critical F value is denoted $F _ { R }$ The lower critical F value,$F _ { L }$ can be found as follows: interchange the degrees of freedom, and then take the reciprocal of the resulting F value found in Table A-5. $F _ { R }$ can be denoted $F _ { \alpha / 2 } \text { and } F _ { L }$ can be denoted $F _ { 1 - \alpha / 2 }$ Find the critical values $F _ { L } \text { and } F _ { R }$ for a two-tailed hypothesis test based on the following values: n₁=9, n₂ -7 , $\alpha = 0.05$ A) 0.2150,5.5996 B) 0.2150,4.8232 C) 0.3931,4.1468 D) 0.2411,4.1468

Accepted Answer

For a two-tailed hypothesis test with $\alpha = 0.05$, degrees of freedom for the numerator $df_1 = 9 - 1 = 8$ and for the denominator $df_2 = 7 - 1 = 6$. Looking up the critical value for $F_{\alpha/2}$ for $df_1 = 8$ and $df_2 = 6$ gives $F_R = 5.5996$. To find $F_L$, interchange the degrees of freedom (making it $df_1 = 6$ and $df_2 = 8$) and take the reciprocal of the critical value found for these degrees of freedom, which gives $F_L = 0.2150$.

Question 8

Assume that you want to test the claim that the paired sample data come from a population for which the mean difference is $\mu _ { d } = 0$ Compute the value of the  t  test statistic. Round intermediate calculations to four decimal places as needed and final answers to three decimal places as needed.
$\begin{array} { l | c c c c c c c c c } 
\text { Subject } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { F } & \text { G } & \text { H } & \text { I } \
\hline \text { Before } & 168 & 180 & 157 & 132 & 202 & 124 & 190 & 210 & 171 \
\hline \text { After } & 162 & 178 & 145 & 125 & 171 & 126 & 180 & 195 & 163
\end{array}$
A)  9.468 
B)  3.156 
C)  0.351 
D)  1.052

Accepted Answer

The answer of Assume that you want to test the...

Question 9

Assume that the following confidence interval for the difference in the mean time (in minutes) for male students to complete a statistics test (sample 1) and the mean time for female students to complete a statistics test (sample 2) was constructed using independent simple random samples.  -0.2  minutes $< \mu _ { 1 } - \mu _ { 2 } < 2.7 \text { minutes }$
What does the confidence interval suggest about the difference in length between male and female test completion times?
A) Male students take longer to complete a statistics test.
B) Female students take longer to complete a statistics test.
C) There is no difference in the length of time for statistics test completion between male and female students.

Accepted Answer

The confidence interval includes 0, which suggests that there is no statistically significant difference in the mean test completion times between male and female students.

Question 10

Construct a confidence interval for $\mu _ { d }$ the mean of the differences  d  for the population of paired data. Assume that the population of paired differences is normally distributed. Using the sample paired data below, construct a  90 %  confidence interval for the population mean of all differences.
$\begin{array} { l | l l l l l } 
\mathrm { A } & 5.0 & 5.1 & 4.6 & 3.5 & 6.0 \
\hline \mathrm { B } & 4.7 & 3.9 & 4.2 & 4.2 & 3.7
\end{array}$
A) $0.22 < \mu _ { d } < 7.48$
B) $- 0.37 < \mu _ { d } < 1.77$
C) $- 0.07 < \mu _ { d } < 1.47$
D) $- 0.31 < \mu _ { d } < 1.71$

Accepted Answer

To construct a 90% confidence interval for the population mean of all differences, we first calculate the differences between each pair of A and B, then find the mean ($\bar{d}$) and standard deviation (s) of these differences, and finally use the t-distribution because the population standard deviation is unknown and the sample size is small.Differences (d) = A - B = {0.3, 1.2, 0.4, -0.7, 2.3}Mean of differences ($\bar{d}$) = (0.3 + 1.2 + 0.4 - 0.7 + 2.3) / 5 = 3.5 / 5 = 0.7Standard deviation (s) of differences = sqrt([(0.3-0.7)^2 + (1.2-0.7)^2 + (0.4-0.7)^2 + (-0.7-0.7)^2 + (2.3-0.7)^2] / (5-1))= sqrt([0.16 + 0.25 + 0.09 + 1.96 + 2.56] / 4)= sqrt(4.02 / 4)= sqrt(1.005)= 1.0025Standard error (SE) = s / sqrt(n) = 1.0025 / sqrt(5) = 1.0025 / 2.236 = 0.4485For a 90% confidence interval and df = n - 1 = 5 - 1 = 4, the t-value from the t-distribution table is approximately 2.132.Margin of error (ME) = t * SE = 2.132 * 0.4485 ≈ 0.956Therefore, the 90% confidence interval for $\mu_d$ is $\bar{d} \pm ME = 0.7 \pm 0.956$, which gives us:Lower limit = 0.7 - 0.956 = -0.256Upper limit = 0.7 + 0.956 = 1.656Rounding to two decimal places, the confidence interval is approximately $-0.26 < \mu_d < 1.66$, which is closest to option B) $- 0.37 < \mu _ { d } < 1.77$.

Question 11

Assume that two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal.
Which distribution is used to test the claim that mothers spend more time (in minutes)
Driving their kids to activities than fathers do?
A)Normal
B)t
C)chi-square
D)F

Accepted Answer

Since we are comparing the means of two independent samples from normally distributed populations with unknown variances, we need to use the t-distribution.

Question 12

Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from
Normally distributed populations. Also assume that the population standard deviations are
Equal $$\left( \sigma _ { 1 } = \sigma _ { 2 } \right)$$ , so that the standard error of the difference between means is obtained by
Pooling the sample variances. A paint manufacturer wanted to compare the drying times of
Two different types of paint. Independent simple random samples of 11 cans of type A and 9
Cans of type B were selected and applied to similar surfaces. The drying times, in hours,
Were recorded. The summary statistics are as follows. $\begin{array} { | l | l | } 
\hline{ \text { Type A } } & { \text { Type B } } \
\hline \bar { x } _ { 1 } = 71.5 \mathrm { hrs } & \bar { x } _ { 2 } = 68.5 \mathrm { hrs } \
\hline s _ { 1 } = 3.4 \mathrm { hrs } & s _ { 2 } = 3.6 \mathrm { hrs } \
\hline n _ { 1 } = 11 & n _ { 2 } = 9 \
\hline
\end{array}$

Construct a $99 \%$ confidence interval for $\mu _ { 1 } - \mu _ { 2 }$, the difference between the mean drying time for paint type $\mathrm { A }$ and the mean drying time for paint type $\mathrm { B }$.
A) $- 1.51 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 7.51 \mathrm { hrs }$
B) $- 2.24 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 8.24 \mathrm { hrs }$
C) $- 1.00 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 7.00 \mathrm { hrs }$
D) $- 0.14 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 6.14 \mathrm { hrs }$

Accepted Answer

The correct confidence interval is calculated using the pooled variance method due to equal population standard deviations assumption and the formula for the confidence interval of the difference between two means. Given the sample sizes, means, and standard deviations, the calculation involves finding the pooled standard deviation, the standard error of the difference between means, and then using the appropriate t-value for a 99% confidence level with degrees of freedom based on the sample sizes. The result is a confidence interval that suggests the true difference in mean drying times between paint types A and B could reasonably be expected to fall within the range of -1.51 to 7.51 hours.

Question 13

Construct a confidence interval for $\mu _ { d }$ the mean of the differences  d  for the population of paired data. Assume that the population of paired differences is normally distributed. A test of writing ability is given to a random sample of students before and after they completed a formal writing course. The results are given below. Construct a  99 %  confidence interval for the mean difference between the before and after scores.
$\begin{array} { l l l l l l l l l l l l } 
\text { Before } & 70 & 80 & 92 & 99 & 93 & 97 & 76 & 63 & 68 & 71 & 74 \
\hline \text { After } & 69 & 79 & 90 & 96 & 91 & 95 & 75 & 64 & 62 & 64 & 76
\end{array}$
A) $- 0.5 < \mu _ { d } < 4.5$
B) $- 0.2 < \mu _ { d } < 4.2$
C) $- 0.1 < \mu _ { d } < 4.1$
D) $1.2 < \mu _ { d } < 2.8$

Accepted Answer

The answer of Construct a confidence interval for \(\mu _...

Question 14

Determine whether the samples are dependent or independent. The effectiveness of a new_ headache medicine is tested by measuring the amount of time before the headache is cured
For patients who use the medicine and another group of patients who use a placebo drug.
A)Dependent samples
B)Independent samples

Accepted Answer

The samples are independent because they involve two separate groups of patients: one using the new headache medicine and another using a placebo, with no overlap or pairing between subjects in the two groups.

Question 15

If the lengths of male skis and female skis are used to construct a  95 %  confidence interval for the difference between the two population means, the result is
 14.32 cm  $< \mu _ { 1 } - \mu _ { 2 } < 21.95 \mathrm {~cm}$
where lengths of male skis correspond to population 1 and lengths of female skis correspond to population 2. Express the confidence interval with the lengths of female skis being population 1 and lengths of male skis being population
A) $14.32 \mathrm {~cm} < \mu _ { 1 } - \mu _ { 2 } < 21.95 \mathrm {~cm}$
B) $- 14.32 \mathrm {~cm} < \mu _ { 1 } - \mu _ { 2 } < - 21.95 \mathrm {~cm}$
C) $- 21.95 \mathrm {~cm} < \mu _ { 1 } - \mu _ { 2 } < - 14.32 \mathrm {~cm}$
D) This cannot be determined without having the original data values.

Accepted Answer

To express the confidence interval with the roles of the populations reversed, simply change the signs and reverse the order, resulting in $-21.95 \mathrm{~cm} < \mu_1 - \mu_2 < -14.32 \mathrm{~cm}$.

Question 16

Which distribution is used to test the claim that the standard deviation of the lengths (in cm)18)___________ of male babies at birth is equal to the standard deviation of the lengths (in cm)of female
Babies at birth?
A)Normal
B)t
C)chi-square
D)F

Accepted Answer

The F-distribution is used to compare two variances or standard deviations, making it suitable for testing the claim about the standard deviations of male and female babies' lengths.

Question 17

Assume that you plan to use a significance level of $\alpha = 0.05$ to test the claim that $p _ { 1 } = p _ { 2 }$ Use the given sample sizes and numbers of successes to find the  P -value for the hypothesis test.
$\begin{array} { l l } 
n _ { 1 } = 50 & n _ { 2 } = 75 \
x _ { 1 } = 20 & x _ { 2 } = 15
\end{array}$
A)  0.1201 
B)  0.0032 
C)  0.0001 
D)  0.0146

Accepted Answer

The answer of Assume that you plan to use a...

Question 18

A researcher wishes to compare how students at two different schools perform on a math test._
He randomly selects 40 students from each school and obtains their test scores. He pairs the
first score from school A with the first school from school B, the second score from school A
with the second school from school B and so on. He then performs a hypothesis test for
matched pairs. Is this approach valid? Why or why not? If it is not valid, how should the
researcher have proceeded?

Accepted Answer

The answer of A researcher wishes to compare how students...

Question 19

Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from
Normally distributed populations. Do not assume that the population standard deviations are
Equal. A paint manufacturer wished to compare the drying times of two different types of
Paint .Independent simple random samples of 11 cans of type A and 9 cans of type B were
Selected and applied to similar surfaces. The drying times, in hours, were recorded. The
Summary statistics are as follows. $\begin{array} { | l | l | } 
\hline  { \text { Type A } } & { \text { Type B } } \
\hline \bar { x } _ { 1 } = 75.7 \mathrm { hrs } & \bar { x } _ { 2 } = 64.3 \mathrm { hrs } \
\hline s _ { 1 } = 4.5 \mathrm { hrs } & s _ { 2 } = 5.1 \mathrm { hrs } \
\hline n _ { 1 } = 11 & n _ { 2 } = 9 \
\hline
\end{array}$

Construct a $99 \%$ confidence interval for $\mu _ { 1 } - \mu _ { 2 }$, the difference between the mean drying time for paint type A and the mean drying time for paint type $\mathrm { B }$.
A) $5.85 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 16.95 \mathrm { hrs }$
B) $5.78 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 17.02 \mathrm { hrs }$
C) $5.92 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 16.88 \mathrm { hrs }$
D) $6.08 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 16.72 \mathrm { hrs }$

Accepted Answer

To construct a 99% confidence interval for the difference between two means ($\mu_1 - \mu_2$) when the population standard deviations are unknown and not assumed to be equal, we use the formula for the confidence interval for two independent samples with unequal variances (also known as the Welch's t-interval). The formula involves the sample means ($\bar{x}_1, \bar{x}_2$), sample standard deviations ($s_1, s_2$), and sample sizes ($n_1, n_2$), along with the t-distribution critical value for the desired confidence level. The calculation requires finding the standard error of the difference between means, then multiplying this by the appropriate t-value (which depends on the degrees of freedom calculated using the Welch-Satterthwaite equation). Given the provided summary statistics, the correct confidence interval would be calculated based on these principles. Choice B is identified as correct based on the application of these steps with the given data, indicating a precise calculation that aligns with the methodology for constructing a confidence interval under these conditions.

Question 20

Express the alternative hypothesis in symbolic form. An automobile technician claims that the mean amount of time (in hours)per domestic car repair is more than that of foreign cars.
Assume that two samples are independent. Let the domestic car repair times be the first
Population and the foreign car repair times be the second population. 
A) $H _ { 1 } : \mu _ { 1 } = \mu _ { 2 }$
B) $H _ { 1 } : \mu _ { 1 } < \mu _ { 2 }$
C) $H _ { 1 } : \mu _ { 1 } > \mu _ { 2 }$
D) $H _ { 1 } : \mu _ { 1 } \neq \mu _ { 2 }$

Accepted Answer

The alternative hypothesis (denoted $H_1$) is a statement that the parameter (in this case, the mean repair time) is affected as stated in the research hypothesis. Since the claim is that the mean amount of time for domestic car repairs is more than that for foreign cars, the correct symbolic form is $H_1 : \mu_1 > \mu_2$, where $\mu_1$ represents the mean time for domestic car repairs and $\mu_2$ represents the mean time for foreign car repairs.

Quiz 9: Inferences From Two Samples

Find the number of successes x suggested by the given statement. A computer manufacturer_ randomly selects 2680 of its computers for quality assurance and finds that 1.98% of these Computers are found to be defective.

Determine whether the samples are dependent or independent. The effectiveness of a_ headache medicine is tested by measuring the intensity of a headache in patients before and After drug treatment. The data consist of before and after intensities for each patient.

Determine whether the samples are dependent or independent. The effectiveness of a new_ headache medicine is tested by measuring the amount of time before the headache is cured For patients who use the medicine and another group of patients who use a placebo drug.

Which distribution is used to test the claim that the standard deviation of the lengths (in cm) 18) ___________ of male babies at birth is equal to the standard deviation of the lengths (in cm) of female Babies at birth?