Question 1

Diet and health have become big business,but many people are not following good health practices.In 1980,a survey of 150 indicated that 30 percent said that at least one time during that year they had five or more alcoholic drinks in one sitting.In 2000,a similar survey of 150 showed that 38 percent had five drinks in one sitting at least one time during the year.Does this represent a significant increase? Use 0.01 level of significance.
A) The test statistic of -1.46 is not less than -1.96; therefore,the sample evidence indicates that there has been no significant change in the proportion of those who drink heavily at least one day during the year.
B) The test statistic of -1.46 is not less than -2.58; therefore,the sample evidence indicates that there has been no significant change in the proportion of those who drink heavily at least one day during the year.
C) The test statistic of -1.46 is not less than -2.33; therefore,the sample evidence indicates that there has not been significant change in the proportion of those who drink heavily at least one day during the year.
D) The test statistic of -2.56 is less than -2.33; therefore,the sample evidence indicates that there has been significant change in the proportion of those who drink heavily at least one day during the year.
E) The test statistic of -2.56 is less than -2.58; therefore,the sample evidence indicates that there has been significant change in the proportion of those who drink heavily at least one day during the year.

Accepted Answer

We need to conduct a two-sample z-test for proportions to determine whether there has been a significant increase in the proportion of people who drink heavily at least one day during the year. Using a significance level of 0.01, the critical value for a two-tailed test is ±2.33. The test statistic is calculated as:

z = (p2 - p1) / sqrt(p * (1 - p) * (1/n1 + 1/n2))

where p1 is the proportion from the first survey, p2 is the proportion from the second survey, p is the pooled proportion, n1 and n2 are the sample sizes from the first and second surveys, respectively.

In this case, p1 = 0.30, p2 = 0.38, n1 = n2 = 150. The pooled proportion is:

p = (x1 + x2) / (n1 + n2) = (30 + 38) / (150 + 150) = 0.34

Plugging in the values, we get:

z = (0.38 - 0.30) / sqrt(0.34 * 0.66 * (1/150 + 1/150)) = 1.46

Since |-1.46| < 2.33, we fail to reject the null hypothesis that there has been no significant change in the proportion of those who drink heavily at least one day during the year. Therefore, the answer is C.

Question 2

When the necessary conditions are met,a two-tail test is being conducted to test the difference between two population proportions.The two sample proportions are p₁ = 0.25 and p₂ = 0.20,and the standard error of the sampling distribution of p₁ - p₂ is 0.04.The calculated value of the test statistic will be: A) z = 0.25. B) z = 1.25. C) t = 0.25. D) t = 0.80.

Accepted Answer

Since the standard error is given, we can use the formula:

test statistic = (sample proportion difference - hypothesized proportion difference) / standard error

Here, the hypothesized proportion difference is 0 (since we are testing for a difference) and the sample proportion difference is 0.25 - 0.20 = 0.05. Plugging in the values, we get:

test statistic = 0.05 / 0.04 = 1.25

Since this is a two-tail test, we need to find the critical values for a significance level (alpha) of 0.05. Looking at a standard normal distribution table or using a calculator, we find that the critical values are approximately -1.96 and 1.96. Since our test statistic (1.25) is not within this range, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest a significant difference between the two population proportions. The correct answer is therefore (B).

Question 3

In testing the difference between two population means using two independent samples,the sampling distribution of the sample mean difference $\overline { \mathrm { x } } _ { 1 } - \overline { \mathrm { x } } _ { 2 }$ is normal if the:
A) sample sizes are both greater than 30.
B) populations are normal.
C) populations are non-normal and the sample sizes are large.
D) population sizes are both greater than 30.
E) all of these are required conditions.

Accepted Answer

The answer of In testing the difference between two population...

Question 4

Chrysler corp
Chrysler Corp.is responding to an advertisement by Toyota,which claims that its compact version gets much better gasoline mileage for in-city driving than any U.S.made car.Below are the results of surveys conducted to test the claim.Perform the appropriate test at the 0.01 significance level. $$\begin{array} { | c | c | } 
\hline \text { Toyota } & \text { Clurysler } \
\hline n _ { 1 } = 30 & n _ { 2 } = 30 \
\bar { x } _ { 1 } = 24 & \bar { x } _ { 2 } = 22.9 \
s _ { 1 } ^ { 2 } = 91.28 & s _ { 2 } ^ { 2 } = 169.3 \
\hline
\end{array}$$
What is the p-value?

Accepted Answer

The answer of Chrysler corp
Chrysler Corp.is responding to an advertisement...

Question 5

Realtor
A new realtor in a large community is attempting to determine differences in the selling prices of houses in two sections of the community.Population 1 is the Northeast and Population 2 is the Southwest.The realtor is going to perform the hypothesis test: HS1U1B10S1U1B0: $\mu _ { 1 } = \mu _ { 2 }$ HS1U1B11S1U1B0: $\mu _ { 1 } \neq \mu _ { 2 }$ A random sample of 35 sale homes is taken from the Northeast and 41 from those in the Southwest using the multiple listing services.A Minitab summary of the results of the two samples expressed in thousands follows: $$\begin{array} { l c c c c c c } 
& \text { N } & \text { MEAN } & \text { MEDIAN } & \text { TRMEAN } & \text { STDEV } & \text { SE MEAN } \
\text { NORTHEAST } & 35 & 86.789 & 84.800 & 88.313 & 60.067 & 10.153 \
\text { SOUTHWEST } & 41 & 60.490 & 62.200 & 61.857 & 14.740 & 2.302
\end{array}$$
-What is the standard error of the difference between the two means?

Accepted Answer

The answer of Realtor
A new realtor in a large community...

Question 6

Price earnings
A comparison of the price-earnings (P/E)ratio for the top and bottom 100 companies in valuation is being prepared.A financial advisor randomly sampled each group to determine whether there is any difference in P/E ratios of the two groups of companies.Let 1 = a top 100 company and 2 = a bottom 100 company.Assume equal population variances and that the populations are normally distributed.The advisor is to use a 0.01 significance level.The data were randomly selected and are summarized below: $$\begin{array} { | c | c | } 
\hline \text { Top 100 company } & \text { Bottom 100 company } \
\hline n _ { 1 } = 6 & n _ { 2 } = 6 \
\bar { x } _ { 1 } = 18.83 & \bar { x } _ { 2 } = 10.67 \
s _ { 1 } ^ { 2 } = 128.17 & s _ { 2 } ^ { 2 } = 8.67 \
\hline
\end{array}$$
Calculate the standard error of the estimate.

Accepted Answer

The answer of Price earnings
A comparison of the price-earnings (P/E)ratio...

Question 7

A political analyst in Texas surveys a random sample of registered Democrats and compares the results with those obtained from a random sample of registered Republicans.This would be an example of:
A) Independent samples.
B) dependent samples.
C) Independent samples only if the sample sizes are equal.
D) Dependent samples only if the sample sizes are equal.

Accepted Answer

The samples of registered Democrats and registered Republicans are two completely separate groups and are not dependent on each other. Therefore, this is an example of independent samples. The size of the samples does not affect whether they are independent or dependent.

Question 8

In testing the difference between two population means using two independent samples,we use the pooled variance in estimating the standard error of the sampling distribution of the sample mean difference $\overline { \mathrm { x } } _ { 1 } - \overline { \mathrm { x } } _ { 2 }$ if the:
A) sample sizes are both large.
B) populations are at least normally distributed with equal variances.
C) populations are non-normal with unequal variances.
D) All of these are required conditions.

Accepted Answer

The pooled variance is used when we have two independent samples from normally distributed populations with equal variances. In this case, the pooled variance is a weighted average of the sample variances from each of the two samples. If the population variances are not equal or the populations are non-normal, then the assumption of normality may not hold and we may need to use non-parametric tests or other procedures. If the sample sizes are both large, we can use the central limit theorem to justify the use of the pooled variance. However, this is not a sufficient condition on its own for using the pooled variance.

Question 9

Which of the following statements are true?
A) The pooled-variances t-test is used whenever the population standard deviations are assumed to be equal regardless of the sample size.
B) The unequal-variances t-test is used whenever the population standard deviations are unknown and cannot be assumed to be equal.
C) The z-test can be used as a close approximation to the unequal-variances t-test when the population standard deviations are not assumed to be equal,but samples are large (each n $\geq$ 30).
D) All of these statements are true.
E) None of these statements is true.

Accepted Answer

A) The pooled-variances t-test is used when the population standard deviations are assumed to be equal, but it can be used regardless of the sample size.
B) The unequal-variances t-test is used when the population standard deviations are unknown and cannot be assumed to be equal.
C) The z-test can be used as a close approximation to the unequal-variances t-test when the population standard deviations are not assumed to be equal, but samples are large (each n ≥ 30).
Therefore, all of these statements are true.

Question 10

A special hypothesis test,based on the t-distribution,can be applied to test the null hypothesis that the population variances are equal for two independent samples.

Accepted Answer

The test based on the t-distribution is used to compare the means of two independent samples, not their variances. To test the equality of variances, tests like the F-test or Levene's test are used.

Question 11

Assume two normally distributed populations with equal population standard deviations.A random sample is taken from each population.Determine the standard error of the estimate for the following situation: $$\begin{array} { l l l l } 
n _ { 1 } & n _ { 2 } & s _ { 1 } & s _ { 2 } \
16 & 14 & 8 & 5
\end{array}$$

Accepted Answer

The answer of Assume two normally distributed populations with equal...

Question 12

Traffic Coordinator A traffic coordinator for a community wishes to determine whether or not there is any difference in the average age of female and male drivers involved in rear-end accidents.The hypothesis test is: H₀: $\mu _ { 1 } = \mu _ { 2 }$ (no difference in mean age) H₁: $\mu _ { 1 } \neq \mu _ { 2 }$ (mean age is different)A random sample of males and females involved in rear-end accidents results in the following Minitab summary: $$\begin{array} { l c c c c c c } & \text { N } & \text { MEAN } & \text { MEDIAN } & \text { TRMEAN } & \text { STDEV } & \text { SEMEAN } \ \text { FEMALES } & 33 & 23.91 & 20.00 & 23.38 & 9.77 & 1.70 \ \text { MALES } & 38 & 28.87 & 28.50 & 28.44 & 9.67 & 1.57 \end{array}$$ What is the standard error of the difference between the two means?

Accepted Answer

The answer of Traffic Coordinator
A traffic coordinator for a community...

Question 13

A sample of size 100 selected from one population has 60 successes,and a sample of size 150 selected from a second population has 95 successes.The test statistic for testing the equality of the population proportions equal to:
A) -0.5319.
B) 0.7293.
C) -0.419.
D) 0.2702.

Accepted Answer

The test statistic for testing the equality of two population proportions is calculated using the formula for the difference between two proportions. The formula involves the sample proportions, the combined proportion, and the sizes of the samples. The correct calculation results in a test statistic of -0.5319.

Question 14

In testing whether the means of two normal populations are equal,summary statistics computed for two independent samples are as follows: $$\begin{array} { | c | c | } 
\hline \text { Brand } \boldsymbol { X } & \text { Brand } \boldsymbol { Y } \
\hline n _ { 1 } = 25 & n _ { 2 } = 25 \
\bar { x } _ { 1 } = 7.30 & \bar { X } _ { 2 } = 6.80 \
s _ { 1 } = 1.05 & s _ { 2 } = 1.20 \
\hline
\end{array}$$ Assume that the population variances are equal.Then,the standard error of the sampling distribution of the sample mean difference $\overline { \mathrm { x } } _ { 1 } - \overline { \mathrm { x } } _ { 2 }$ is equal to:
A) 0.1017.
B) 1.2713.
C) 0.3189.
D) 1.1275.

Accepted Answer

The standard error of the difference between two means is calculated using the formula: $ SE = \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2}} $, where $ s_p^2 $ is the pooled variance. Here, $ s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2} = \frac{24 \times 1.05^2 + 24 \times 1.20^2}{48} = 1.1275 $. Thus, $ SE = \sqrt{\frac{1.1275}{25} + \frac{1.1275}{25}} = 0.3189 $.

Question 15

Women drivers
In the past,young women drivers have maintained a better driving record than young men drivers.An insurance company is concerned with the driving record of its insured customers.Specifically,it conducts a test for the number of speeding tickets received during the past year by drivers between the ages of 18 and 25. $$\begin{array} { | c | c | } 
\hline \text { Men } & \text { Women } \
\hline n _ { 1 } = 120 & n _ { 2 } = 85 \
\bar { x } _ { 1 } = 1.2 & \bar { x } _ { 2 } = 0.4 \
s _ { 1 } ^ { 2 } = 24.8 & s _ { 2 } ^ { 2 } = 10.6 \
\hline
\end{array}$$
A test for the equality of average number of tickets per driver for the two groups is desired.Use $\alpha$ = 0.01.Calculate the p-value for this test.

Accepted Answer

The answer of Women drivers
In the past,young women drivers have...

Question 16

When the necessary conditions are met,a two-tail test is being conducted to test the difference between two population proportions.If the value of the test statistic z is 2.05,then the p-value is:
A) 0.4798.
B) 0.0404.
C) 0.2399.
D) 0.0202.

Accepted Answer

Since it is a two-tail test, we need to find the area in both tails of the standard normal distribution. A z-score of 2.05 indicates that the test statistic is 2.05 standard deviations away from the mean. Looking at the standard normal distribution table, the area in the tail to the right of 2.05 is 0.0202. However, we need to find the total area in both tails, so we need to double this value to get the p-value, which is 0.0404. Therefore, the answer is B.

Question 17

When the necessary conditions are met,a two-tail test is being conducted to test the difference between two population means,but your statistical software provides only a one-tail area of 0.036 as part of its output.The p-value for this two-tail test will be:
A) 0.018.
B) 0.009.
C) 0.072.
D) 0.964.

Accepted Answer

In a two-tail test, the p-value from a one-tail test is doubled to account for both tails of the distribution. Therefore, if the one-tail area (p-value) is 0.036, the two-tail p-value would be 0.036 * 2 = 0.072.

Question 18

Realtor
A new realtor in a large community is attempting to determine differences in the selling prices of houses in two sections of the community.Population 1 is the Northeast and Population 2 is the Southwest.The realtor is going to perform the hypothesis test: HS1U1B10S1U1B0: $\mu _ { 1 } = \mu _ { 2 }$ HS1U1B11S1U1B0: $\mu _ { 1 } \neq \mu _ { 2 }$ A random sample of 35 sale homes is taken from the Northeast and 41 from those in the Southwest using the multiple listing services.A Minitab summary of the results of the two samples expressed in thousands follows: $$\begin{array} { l c c c c c c } 
& \text { N } & \text { MEAN } & \text { MEDIAN } & \text { TRMEAN } & \text { STDEV } & \text { SE MEAN } \
\text { NORTHEAST } & 35 & 86.789 & 84.800 & 88.313 & 60.067 & 10.153 \
\text { SOUTHWEST } & 41 & 60.490 & 62.200 & 61.857 & 14.740 & 2.302
\end{array}$$
-What is the value of the test statistic?
z = ____________________

Accepted Answer

The answer of Realtor
A new realtor in a large community...

Question 19

In testing for the equality of two population variances,when the populations are normally distributed,the 10% level of significance has been used.To determine the rejection region,it will be necessary to refer to the F table corresponding to an upper-tail area of:
A) 0.90.
B) 0.05.
C) 0.20.
D) 0.10.

Accepted Answer

When testing for the equality of two population variances, we use the F test. In this case, we are using a two-tailed test at the 10% level of significance. This means that we split the alpha (0.10) into two equal tails, each with an area of 0.05.

To determine the rejection region, we need to look up the critical value for the upper tail area of 0.05 in the F table. This critical value corresponds to the degrees of freedom for the numerator (population with the larger variance) and denominator (population with the smaller variance). The numerator degrees of freedom is calculated as the sample size of population 1 minus 1, and the denominator degrees of freedom is calculated as the sample size of population 2 minus 1.

Since the question does not provide any information about the sample sizes of the populations, we cannot calculate the degrees of freedom or look up the critical value. However, we know that the correct answer is B because the upper-tail area of 0.05 is the one that corresponds to the significance level of 0.10 in a two-tailed test. The other answer choices are incorrect and would lead to the wrong rejection region.

Question 20

In testing the difference between two population means using two independent samples,the population standard deviations are assumed to be unknown,each sample size is ? 30,and the calculated test statistic z = 2.56.If the test is two-tail and 5% level of significance has been specified,the conclusion should be to:
A) reject the null hypothesis.
B) not to reject the null hypothesis.
C) choose two other independent samples.
D) none of these answers is correct.

Accepted Answer

Since the calculated test statistic z is greater than the critical value of 1.96 (for a two-tail test at 5% level of significance), we reject the null hypothesis. Thus, the conclusion is that there is a significant difference between the means of two populations.

Quiz 11: Hypothesis Tests Involving Two Sample Means

In testing the difference between two population means using two independent samples,the sampling distribution of the sample mean difference x‾1−x‾2\overline { \mathrm { x } } _ { 1 } - \overline { \mathrm { x } } _ { 2 }x1​−x2​ is normal if the:

A political analyst in Texas surveys a random sample of registered Democrats and compares the results with those obtained from a random sample of registered Republicans.This would be an example of:

Which of the following statements are true?

A special hypothesis test,based on the t-distribution,can be applied to test the null hypothesis that the population variances are equal for two independent samples.

A sample of size 100 selected from one population has 60 successes,and a sample of size 150 selected from a second population has 95 successes.The test statistic for testing the equality of the population proportions equal to:

When the necessary conditions are met,a two-tail test is being conducted to test the difference between two population proportions.If the value of the test statistic z is 2.05,then the p-value is:

When the necessary conditions are met,a two-tail test is being conducted to test the difference between two population means,but your statistical software provides only a one-tail area of 0.036 as part of its output.The p-value for this two-tail test will be:

In testing for the equality of two population variances,when the populations are normally distributed,the 10% level of significance has been used.To determine the rejection region,it will be necessary to refer to the F table corresponding to an upper-tail area of:

In testing the difference between two population means using two independent samples,the sampling distribution of the sample mean difference $\overline { \mathrm { x } } _ { 1 } - \overline { \mathrm { x } } _ { 2 }$ is normal if the: