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(1) Suppose you are interested in comparing by sex the mean number of clinician

ID: 3060983 • Letter: #

Question

(1) Suppose you are interested in comparing by sex the mean number of clinician visits at your agency per year. You hypothesize that women are more likely to visit the clinicians at your agency than men.

(a) Using the data in the table below, conduct a test of your hypothesis using the 5 steps of hypothesis testing. The value of t = -2.038. Use alpha = .05. Cite all statistics [including the critical value of t and your p value] and use words to explain your findings. Points will be deducted for not using all steps of hypothesis testing and providing a full explanation of your results.

Interpret your findings at alpha = .01 Cite your critical value and your p-value.

Visits Men

Range

59

Minimum value

11

Maximum value

70

Mean

37.28

Variance

142.88

Standard deviation

11.95

Sum

2013

Number of observations

54

Visits Women

Range

60

Minimum value

5

Maximum value

65

Mean

42.31

Variance

100.00

Standard deviation

10.00

Sum

1,227

Number of observations

29

(2) Imagine you wish to determine whether the mean number of hours worked per week by adults in a sample that you obtained from your field placement differs from the 40-hour standard that is practiced nationally. Conduct a test of your hypothesis using all the 5 hypothesis steps. Use alpha = .05. Cite all statistics [including the critical value of t and the p value] and use words to explain your findings. Points will be deducted for not using all steps of hypothesis testing and providing a full explanation of your results.

Weekly Hours Worked at Your Agency

Range

60

Minimum value

5

Maximum value

65

Mean

42.31

Variance

100.00

Standard deviation

10.00

Sum

1,227

Number of observations

29

Visits Men

Range

59

Minimum value

11

Maximum value

70

Mean

37.28

Variance

142.88

Standard deviation

11.95

Sum

2013

Number of observations

54

Visits Women

Range

60

Minimum value

5

Maximum value

65

Mean

42.31

Variance

100.00

Standard deviation

10.00

Sum

1,227

Number of observations

29

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: = 40
Alternative hypothesis: 40

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 1.85695

DF = n - 1 = 50 - 1

D.F = 28
t = (x - ) / SE

t = - 0.4289

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 28 degrees of freedom is less than -0.4289 or greater than 0.4289.

Thus, the P-value = 0.67.

Interpret results. Since the P-value (0.67) is greater than the significance level (0.05), we cannot reject the null hypothesis.