Home Chapter 10 Online Study Guide Multiple Choice

# Multiple Choice

This activity contains 20 questions.

## When testing the null hypothesis using the confidence interval estimate of the difference between two means, one would reject the null hypothesis when

 the confidence interval does not include zero the lower confidence limit is less than zero the upper confidence limit is greater than zero the confidence interval includes zero

## When the population standard deviations are unknown, both samples are less than 30, and the equal variances assumption cannot be met, which test statistic should be used to test the differences between two independent means

 pooled-variance t test for the difference between two means the paired t test Z test for the difference between two means separate-variance t-test for the difference between two means

## The t test for the difference between the means of two independent samples assumes that the respective:

 sample variances are equal. populations are approximately normal. all of the above. samples are randomly and independently drawn.

## If we are testing for the difference between the means of two independent samples with samples of n1 = 20 and n2 = 20, the number of degrees of freedom is equal to:

 10 38 39 40

## In testing for the differences between the means of two independent populations where the variances in each population are unknown but assumed equal, the degrees of freedom are:

 n1 + n2 - 1. n - 1. n1 + n2 - 2. n - 2.

## In testing for differences between the means of two independent populations the null hypothesis states that:

 the difference between the two population means is not significantly different from zero. the difference between the two population means is significantly greater than zero. the difference between the two population means is significantly less than zero. the difference between the two population means is significantly different from zero.

## In testing for differences between the means of two related populations (i.e., Matched-Pairs) where the variance of the differences is unknown, the degrees of freedom are:

 n1 + n2 - 2. n1 + n2 - 1. n - 2. n - 1.

## If we are testing for the difference between the means of two related samples with samples of n1 = 20 and n2 = 20, the number of degrees of freedom is equal to:

 38 19 18 39

## When testing for differences between the means of two related populations, the null hypothesis states that:

 the difference between the two population means is not significantly different from zero. the population mean difference is not significantly different from zero. the difference between the two population means is significantly greater than zero. the population mean difference is significantly greater than zero.

## The probability distribution used to test for differences between two population variances is the

 F distribution normal probability distribution t distribution binomial probability distribution

## When testing for the difference between two population variances with sample sizes of n1 = 8 and n2 = 10, the number of degrees of freedom are:

 numerator d.f. =8, denominator d.f. = 10 d.f. = 16 d.f. = 18 numerator d.f. = 7, denominator d.f. = 9

## The hypothesis test for the equality of two population variances is based on:

 the difference between the sample variances divided by the difference between the sample means. the ratio of the two sample variances. the difference between the two sample variances. the difference between the two population variances.

## A Pooled-Variance t Test for the Difference Between Two Independent Means may be used when

 the two population variances are found to be significantly different the sample sizes are equal. the two population variances are found not to be significantly different using a two-tailed test

## Refer to Figure 10.3 on page 373, “Microsoft Excel t test results for the two display locations. Based on the results presented, the pooled-variance would be equal to

 350.678 254.006 157.333 508.011

## Refer to Figure 10.3 on page 373, “Microsoft Excel t test results for the two display locations. Based on the results presented, the computed t test statistic is

 2.1009 18 –3.0446 1.7341

## Refer to Figure 10.3 on page 373, “Microsoft Excel t test results for the two display locations. Based on the results presented, the correct degrees of freedom for this test would be

 19 18 20 9

## Refer to Figure 10.3 on page 373, “Microsoft Excel t test results for the two display locations. Based on the results presented, at what level of statistical significance would the null hypothesis be rejected?

 using a level of significance of five percent using a level of significance of one percent the null hypothesis would not be rejected at the one percent level of significance the null hypothesis would not be rejected at the five percent level of significance

To investigate the efficacy of a diet, a random sample of 16 male patients is drawn from a population of adult male volunteers is collected. The weight of each individual in the sample is taken at the start of the diet, and at a medical follow-up four weeks later. Assume that the population of differences in weight before versus after the diet follow a normal distribution. What would be the appropriate statistical test to conduct the hypothesis test?

 Z test for the difference between two means Paired t test Z test for the difference between two proportions Pooled-Variance t Test for the difference between two means

To investigate the efficacy of a diet, a random sample of 16 male patients is drawn from a population of adult male volunteers is collected. The weight of each individual in the sample is taken at the start of the diet, and at a medical follow-up four weeks later. Assume that the population of differences in weight before versus after the diet follow a normal distribution. If it is hypothesized that the diet will lead to a statistically significant loss in weight, what type of hypothesis test should be conducted?

 neither one- nor two one either one- or two two

See Section 10.3, “Comparing Two Population Proportions,” and refer to Figure #10.11, “Microsoft Excel results for the Z test between two proportions for the hotel guest satisfaction problem,” p. 393. The value for the pooled estimate of the population proportion of successes is

 0.5878 0.0026 0.6483 0.7181