Question 1

You should use a pooled t-test versus a non-pooled t-test when
A) the samples are matched.
B) the variance of the two populations are equal.
C) the sample sizes of both groups are the same.
D) None of the above.
E) A and C.

Accepted Answer

A pooled t-test assumes that the variances of the two populations being compared are equal. Therefore, if the variances are known to be equal or suspected to be equal, a pooled t-test should be used. If the variances are not equal, a non-pooled (also known as an unpooled or separate variances) t-test should be used. The other options (A and C) do not determine whether a pooled or non-pooled t-test should be used.

Question 2

An army depot that overhauls ground mobile radar systems is interested in improving its processes. One problem involves troubleshooting a particular component that has a
High failure rate after it has been repaired and reinstalled in the system. The shop floor
Supervisor believes that having standard work procedures in place will reduce the time
Required for troubleshooting this component. Time (in minutes) required troubleshooting
This component without and with the standard work procedure is recorded for a sample of
19 employees. Assuming that we define our differences as Time without standard work
Procedure - Time with standard work procedure, the correct alternative hypothesis is
A) µd = 0.
B) µd < 0.
C) µd > 0.
D) µd ≠ 0.
E) None of the above.

Accepted Answer

Since the problem involves reducing the time required for troubleshooting, we want to see if the standard work procedure has a positive effect (i.e. reduces the time). Therefore, we are interested in whether the mean difference (Time without standard work procedure - Time with standard work procedure) is greater than 0. This leads us to the alternative hypothesis of µd > 0.

Question 3

Data were collected on annual personal time (in hours) taken by a random sample of 16 women and 7 men employed by a medium sized company. The women took an
Average of 24.75 hours of personal time per year with a standard deviation of 2.84 hours.
The men took an average of 21.89 hours of personal time per year with a standard
Deviation of 3.29 hours. The Human Resources Department believes that women tend to
Take more personal time than men because they tend to be the primary child care givers in
The family. The t-test for two means is appropriate in this situation because
A) women and men are dependent samples.
B) women and men are independent samples.
C) women and men are matched samples.
D) the observations are paired.
E) None of the above.

Accepted Answer

Women and men are independent samples because they are not related or matched in any way (i.e., they are not dependent). Therefore, the t-test for two means is appropriate in this situation.

Question 4

A real estate agency, located in a metropolitan area in the northeastern U.S., kept data on the various types of properties purchased in the area. Historically, 15% of purchases
Were for condominiums, 30% were for townhouses, 40% for single family homes, 10%
For commercial properties and 5% for land. Based on the data shown below, the null
Hypothesis was rejected and the agency concluded that the current distribution of property
Sales differs from the historical distribution. Based on an examination of the standardized
Residuals, which type of property had the most impact on the test results? $$\begin{array} { l | c | c | c | c | c } 
\text { Type of Property } & \text { Condos } & \text { Townhouses } & \text { Homes } & \text { Commercial } & \text { Land } \
\hline \text { Frequency } & 89 & 121 & 78 & 25 & 12
\end{array}$$
A) Condos
B) Townhouses
C) Homes
D) Commercial
E) Land

Accepted Answer

The standardized residual for condos is the largest in absolute value, indicating that the difference between the observed and expected frequencies for condos is the most extreme.

Question 5

Which of the following is not an assumption or condition that needs to be checked for a two-sample t-test for the difference between two means?
A) Independent Groups
B) Randomization
C) 10% Condition
D) Nearly Normal Condition
E) All of the above must be checked.

Accepted Answer

All of the choices listed (Independent Groups, Randomization, 10% Condition, Nearly Normal Condition) are assumptions or conditions that need to be checked when conducting a two-sample t-test for the difference between two means.

Question 6

A contact lens wearer read that the producer of a new contact lens boasts that their hypothesis $H _ { 0 } : \mu _ { \text {old } } - \mu _ { \text {new } } = 0$ is tested against the alternative $H _ { A } : \mu _ { \text {old } } - \mu _ { \text {new } } > 0$. Which of the following would be a Type II error?
A) Deciding that the new lenses are cheaper, when in fact they really are.
B) Deciding that the new lenses are cheaper, when in fact they are not.
C) Deciding that the new lenses are not really cheaper, when in fact they are.
D) Deciding that the new lenses are not really cheaper, when in fact they are not.
E) It does not matter.

Accepted Answer

A Type II error is failing to reject a false null hypothesis. In this case, it means deciding that the means of the old and new lenses are the same (not really cheaper), when in fact the new lenses are cheaper (the alternative hypothesis is true). So, option C fits this definition.

Question 7

A type II error is committed when
A) we don't reject a null hypothesis that is true.
B) we reject a null hypothesis that is false.
C) we reject a null hypothesis that is true.
D) we don't reject a null hypothesis that is false.
E) none of the above

Accepted Answer

A type II error is committed when we fail to reject a null hypothesis that is actually false, so choice D is the correct answer.

Question 8

A local politician was interested in determining whether income level affects opinion regarding the government's bailout bill for easing the 2008 financial crisis on Wall Street.
He surveyed a sample of his constituents and got the following results. The calculated
Value of the Chi Square statistic is 
A) 7.433
B) 29.13
C) 40.70
D) 5.999
E) 2.347

Accepted Answer

The answer of A local politician was interested in determining...

Question 9

A real estate agency, located in a metropolitan area in the northeastern U.S., kept data on the various types of properties purchased in the area. Historically, 15% of purchases
Were for condominiums, 30% were for townhouses, 40% for single family homes, 10%
For commercial properties and 5% for land. With changing demographics, the agency
Wondered if the current distribution matches the historical distribution. Recent data
Showed the following: $$\begin{array} { l | c | c | c | c | c } 
\text { Type of Property } & \text { Condos } & \text { Townhouses } & \text { Homes } & \text { Commercial } & \text { Land } \
\hline \text { Frequency } & 89 & 121 & 78 & 25 & 12
\end{array}$$ What is the standardized residual associated with Townhouses?
A) 23.5
B) 4.85
C) 10.86
D) 2.38
E) Not enough information given to calculate.

Accepted Answer

To calculate the standardized residual, we need to first calculate the expected frequency for each type of property based on the historical distribution. We can do this by multiplying the total number of purchases (325) by the historical percentage for each type of property.

Condos: 0.15 x 325 = 48.75
Townhouses: 0.30 x 325 = 97.5
Homes: 0.40 x 325 = 130
Commercial: 0.10 x 325 = 32.5
Land: 0.05 x 325 = 16.25

We can then calculate the expected frequency for townhouses by subtracting the historical percentage from the observed percentage and multiplying by the total number of purchases.

Expected frequency for townhouses = (0.30 - 121/325) x 325 = 92.65

To calculate the standardized residual for townhouses, we use the formula:

Standardized residual = (observed frequency - expected frequency) / square root(expected frequency)

For townhouses, the observed frequency is 121 and the expected frequency is 92.65. Plugging these values into the formula, we get:

Standardized residual = (121 - 92.65) / square root(92.65) = 2.38

Therefore, the answer is D) 2.38.

Question 10

An army depot that overhauls ground mobile radar systems is interested in improving its processes. One problem involves troubleshooting a particular component that has a
High failure rate after it has been repaired and reinstalled in the system. The shop floor
Supervisor believes that having standard work procedures in place will reduce the time
Required for troubleshooting this component. Time (in minutes) required troubleshooting
This component without and with the standard work procedure is recorded for a sample of
19 employees. In order to determine if having a standard work procedure in place
Reduces troubleshooting time, they should use
A) a one-tailed paired t-test.
B) a two-tailed test of two independent means.
C) a one-tailed test of two independent means.
D) a two-tailed paired t-test.
E) a test of two proportions.

Accepted Answer

A one-tailed paired t-test is appropriate because the data involves comparing the times before and after implementing a standard work procedure for the same group of employees, indicating paired data, and the hypothesis specifically looks for a reduction (not just any difference) in time, making it one-tailed.

Question 11

In economic downturns, companies attempt to downsize their workforces by offering early retirement incentives to older employees. A survey of 723 companies found that
195 engage in such downsizing practices. The 99% confidence interval for the
Proportion of companies that downsize their workforces by offering early retirement
Incentives is
A) 0.19 to 0.35
B) 0.65 to 0.81
C) 0.19 to 0.47
D) 0.69 to 0.77
E) 0.23 to 0.31

Accepted Answer

The proportion of companies offering early retirement incentives is 195/723. Using this proportion, a 99% confidence interval is calculated, resulting in the interval 0.23 to 0.31.

Question 12

A real estate agency, located in a metropolitan area in the northeastern U.S., kept data on the various types of properties purchased in the area. Historically, 15% of
Purchases were for condominiums, 30% were for townhouses, 40% for single family
Homes, 10% for commercial properties and 5% for land. With changing demographics,
The agency wondered if the current distribution matches the historical distribution.
Recent data showed the following: $$\begin{array} { l | c | c | c | c | c } 
\text { Type of Property } & \text { Condos } & \text { Townhouses } & \text { Homes } & \text { Commercial } & \text { Land } \
\hline \text { Frequency } & 89 & 121 & 78 & 25 & 12
\end{array}$$ Which Chi-square test is most appropriate for this situation?
A) Goodness of Fit
B) Homogeneity
C) Independence
D) Proportional
E) None of the above.

Accepted Answer

The Goodness of Fit test is used to test whether an observed frequency distribution fits a hypothesized distribution. In this case, the agency wants to test whether the observed frequency distribution of property purchases matches the historical distribution. Therefore, the most appropriate test is the Goodness of Fit test.

Question 13

A local politician was interested in determining whether income level affects opinion regarding the government's bailout bill for easing the 2008 financial crisis on Wall Street.
He surveyed a sample of his constituents and got the following results. The correct null
Hypothesis to be tested is 
A) There is no relationship between income level and opinion.
B) There is a relationship between income level and opinion.
C) Opinion and income level are independent.
D) Both A and C
E) Both B and C

Accepted Answer

The answer of A local politician was interested in determining...

Question 14

Which factor listed below does not affect the power of a test?
A) Sample size
B) Standard deviation
C) Significance level
D) One-sided versus two-sided tests
E) All of the above

Accepted Answer

All the options listed (A: Sample size, B: Standard deviation, C: Significance level, D: One-sided versus two-sided tests) are factors that affect the power of a statistical test.

Question 15

A real estate agency, located in a metropolitan area in the northeastern U.S., kept data on the various types of properties purchased in the area. Historically, 15% of purchases
Were for condominiums, 30% were for townhouses, 40% for single family homes, 10%
For commercial properties and 5% for land. With changing demographics, the agency
Wondered if the current distribution matches the historical distribution. Recent data
Showed the following: $$\begin{array} { l | c | c | c | c | c } 
\text { Type of Property } & \text { Condos } & \text { Townhouses } & \text { Homes } & \text { Commercial } & \text { Land } \
\hline \text { Frequency } & 89 & 121 & 78 & 25 & 12
\end{array}$$ The value of the Chi-square test statistic is
A) χ 2 = 5.993
B) χ 2 = 62.538
C) χ 2 = 101.482
D) χ 2 = 77.431
E) χ 2 = 8.101

Accepted Answer

The Chi-square test statistic is calculated by comparing the observed frequencies with the expected frequencies based on historical percentages. The calculation results in a test statistic of 62.538.

Question 16

A local politician was interested in determining whether income level affects opinion regarding the government's bailout bill for easing the 2008 financial crisis on Wall Street.
He surveyed a sample of his constituents and performed a Chi-square test. The resulting
P-value was 0.024. At the 0.01 level of significance, he should conclude that
A) Income level affects opinion.
B) There is a relationship between income level and opinion.
C) Opinion and income level are independent.
D) Reject the null hypothesis.
E) Both B and D.

Accepted Answer

The P-value of 0.024 is greater than the significance level of 0.01, so we fail to reject the null hypothesis, which suggests that opinion and income level are independent.

Question 17

Shape-Up-By-Susie claims that participating in her exercise program will result in guaranteed weight loss in just 6 weeks. Five clients weighed themselves before and then
Again after participating in her program and the differences "weight after - weight
Before" were analyzed. Based on the relevant computer output shown below, we can say Paired T for After - Before
$\begin{array} { l c r r r } & \mathrm { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \ \text { After } & 5 & 144.6 & 31.7 & 14.2 \ \text { Before } & 5 & 148.0 & 28.9 & 12.9 \ \text { Difference } & 5 & - 3.40 & 3.51 & 1.57 \end{array}$
$95 \%$ upper bound for mean difference: $- 0.06$
$\mathrm { T }$-Test of mean difference $= 0 ( \mathrm { vs } < 0 ) : \mathrm { T }$-Value $= - 2.17 \mathrm { P }$-Value $= 0.048$
A) This is a paired design.
B) This is a two tailed test.
C) This is one tailed test.
D) Both A and B.
E) Both A and C.

Accepted Answer

This is a paired design because the same subjects were weighed before and after the exercise program. 
Since the null hypothesis is that there is no difference between the mean weights before and after the program (i.e. mean difference = 0), and the alternative hypothesis is that the mean weight after the program is less than the mean weight before the program (one-sided), this is a one-tailed test.

Question 18

Shape-Up-By-Susie claims that participating in her exercise program will result in guaranteed weight loss in just 6 weeks. Five clients weighed themselves before and then
Again after participating in her program and the differences "weight after - weight
Before" were analyzed. The relevant output is shown below. At the 0.05 level of
Significance, the correct conclusion is Paired $T$ for After - Before
$\begin{array} { l c r r r } & \mathrm { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \ \text { After } & 5 & 144.6 & 31.7 & 14.2 \ \text { Before } & 5 & 148.0 & 28.9 & 12.9 \ \text { Difference } & 5 & - 3.40 & 3.51 & 1.57 \end{array}$
$95 \div$ upper bound for mean difference: $- 0.06$
T-Test of mean difference $= 0 ( \mathrm { vs } < 0 ) : \mathrm { T }$-Value $= - 2.17$ P-Value $= 0.048$
A) Reject the null hypothesis.
B) Do not reject the null hypothesis.
C) Susie's claim is supported by the sample evidence.
D) Both A and C.
E) Both B and C.

Accepted Answer

The null hypothesis in this case is that there is no significant difference between the weights before and after participating in Susie's exercise program. The alternative hypothesis is that there is a significant difference, which would support Susie's claim of guaranteed weight loss.

The calculated T-value of -2.17 is less than the critical T-value of -2.776 (based on degrees of freedom = 4 and alpha level = 0.05), indicating that we should reject the null hypothesis.

Therefore, we can conclude that there is evidence to support Susie's claim of guaranteed weight loss in just 6 weeks based on the sample data.

Question 19

After computing a confidence interval, the investigator believes that the results are meaningless because the width of the interval is too large. In reconstructing the interval,
The investigator should
A) decrease the sample size
B) increase the level of confidence
C) increase the sample size
D) reduce the population variance
E) none of the above

Accepted Answer

Increasing the sample size will reduce the width of the confidence interval. This is because as sample size increases, the standard error of the estimate decreases. Decreasing the sample size will likely increase the width of the confidence interval. Increasing the level of confidence or reducing the population variance may have some effect on the width of the interval, but these measures would not be as effective as increasing the sample size.

Question 20

A human resources manager at a large company wants to estimate the proportion of employees that would be interested in reimbursement for college courses. If she wishes
To be 95% confident that her estimate is within 5% of the true proportion, how many
Employees would need to be sampled?
A) 271
B) 385
C) 543
D) 646
E) 1234

Accepted Answer

To achieve a 95% confidence level with a 5% margin of error when the population proportion is unknown, the formula to calculate the sample size is $n = \frac{Z^2 \times p(1-p)}{E^2}$, where $Z$ is the Z-score corresponding to the confidence level (1.96 for 95%), $p$ is the estimated proportion (0.5 is used when the proportion is unknown for maximum sample size), and $E$ is the margin of error (0.05 for 5%). Substituting these values gives $n = \frac{(1.96)^2 \times 0.5(1-0.5)}{(0.05)^2} = 384.16$. Rounding up to the nearest whole number gives 385.

Quiz 13: Inference for Counts: Chi-Square Tests

You should use a pooled t-test versus a non-pooled t-test when

Which of the following is not an assumption or condition that needs to be checked for a two-sample t-test for the difference between two means?

A type II error is committed when

A local politician was interested in determining whether income level affects opinion regarding the government's bailout bill for easing the 2008 financial crisis on Wall Street. He surveyed a sample of his constituents and got the following results. The correct null Hypothesis to be tested is

Which factor listed below does not affect the power of a test?

After computing a confidence interval, the investigator believes that the results are meaningless because the width of the interval is too large. In reconstructing the interval, The investigator should

A human resources manager at a large company wants to estimate the proportion of employees that would be interested in reimbursement for college courses. If she wishes To be 95% confident that her estimate is within 5% of the true proportion, how many Employees would need to be sampled?