Question

The following random sample of weekly student expenses in dollars is obtained from a normally distributed population of undergraduate students with unknown parameters.

8	56	76	75	62	81	72	69	91	84
49	75	69	59	70	53	65	78	71	87
71	74	69	65	64

You have been charged to conduct a statistical test in SPSS to verify the claim that the‘average weekly student expenses’ is different than 74 dollars using an alpha level of 5%.

           

What is the appropriate test that is applicable in this case. Explain your reasoning.

State the null and alternate hypotheses in this case using proper statistical notations.

List one assumption that you are making about the distribution.

Insert a copy of the summary table of descriptive statistics generated in SPSS.

Insert a copy of the table for the statistical test you conducted in SPSS.

Drawing on information from the tables in (e) and/or (f) show how they relate to t-statistic as obtained in SPSS.

What is/are the critical value(s) of the test statistic at the 5% significance level.

What can you conclude about the claim based on the results generated from the statistical test? Make sure to support your conclusion by referencing the appropriate statistics from the test.

Compute the 90% confidence interval for the average weekly expenses.

Compute the Cohen’s d effect size.

Answer 1

The appropriate test in this case would be one sample t test. Because we have one sample here and we need to test that the‘average weekly student expenses’ is different than 74 dollars .

NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS Ha:

alpha= 0.05

Degrees of freedom= n-1= 25-1=24

The one assumption I'm making here is that the parent population from which sample is taken is normally distributed with unknown standard deviation.

t=

t=-1.964

Since test is two tailed we have

critical value of t for two tailed test is 2.06

Conclusion: Since critical value of t is greater than calculated value of t we therefore do not reject null hypothesis and conclude that we do not have enough evidence to show that the‘average weekly student expenses’ is different than 74 dollars.

90% confidence interval is

Sample Mean = 67.72
t critical = 1.71
sM = √(15.98572/25) = 3.2

μ = M ± t(sM)
μ = 67.72 ± 1.71*3.2
μ = 67.72 ± 5.4699

Result

90% CI [62.2501, 73.1899].

You can be 90% confident that the population mean (μ) falls between 62.2501 and 73.1899.

Cohen's d = (67.72 - 74) ⁄ 15.9857 = 0.392851.

Interpreting Cohen's d

d = 0.2 Small effect - mean difference is 0.2 standard deviation
d = 0.5 Medium effect - mean difference is 0.5 standard deviation
d = 0.8 Large effect - mean difference is 0.8 standard deviation