Question

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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

Since the problem given, is to verify the claim that the average weekly student expenses’ is different from 74 dollars, we use ONE SAMPLE T-TEST.

The one-sample t-test is used to determine whether a sample comes from a population with a specific mean. This population mean is not always known, but is sometimes hypothesized.

SUMMARY TABLE SHOWING DESCRIPTIVE STATISTICS:

For a one-sample t-test, the assumptions are:

a) Independent and identically distributed variables (or “independent observations”)

b) Normality: The test variable is normally distributed in the population.

Before we actually conduct the one sample t-test, our first step is to check the distribution for normality. This is best done with a Q-Q Plot and Kolmogrov-smirnov test.

KOLMOGROV SMIRNOV TEST:

The hypothesis for KS test is given by:

H0:	The data follows a NORMAL distribution.
Ha:	The data do not follow NORMAL distribution.

From the K-S Test, we could see that it is not significant as the pvalue 0.200 is greater than 0.05. Thus we cannot reject H₀ and we conclude that the variable weekly expense is normally distributed.

NORMAL Q-Q PLOT:

From the normal Q-Q plot, it is evident that the distribution of student weekly expense data is normally distributed.

Let’s move on to the one sample t-test:

The hypothesis for one sample t-test is given by,

: The average weekly student expenses’ is equal to 74 dollars.

​: The average weekly student expenses’ is not equal to 74 dollars.

LEVEL OF SIGNIFICANCE:

TEST STATISTIC:

where is sample mean.

​ is the hypothized value 74 .

is sample standard deviation.

is the total sample size.

CRITICAL VALUE:

The two-tailed critical​ t-value with degrees of freedom (n-1 = 24) at ​ is .

DECISION RULE:

If the calculated t-statistic value is greater than t critical value at we reject .

And if p-value is less than or equal to , we reject .

ONE SAMPLE T-TEST RESULTS:

From the one sample t-test output, the t-statistic value (-1.93) is lesser than the critical t-value at 5% level of significance (2.06),also, p-value for one sample t-test is 0.065 which is greater than , we fail to reject null hypothesis and conclude that the average weekly student expenses’ is not equal to 74 dollars.

TABLE DISPLAYING 90% CONFIDENCE INTERVAL FOR AVERAGE WEEKLY EXPENSES:

The 90% confidence interval for the average weekly expenses is given by ​.

COHENS D EFFECT SIZE:

Cohen's d is an effect size used to indicate the standardised difference between two means. A cohen's d effect size near 0.2 is a small effect, a cohen's d effect size near 0.5 is a medium effect, and a cohen's d effect size near 0.8 is a large effect.

Cohen’s d must be calculated by hand, but you can get both of the values used in the formula from the SPSS output:

The standard effect size for a one-sample t-test is the difference between the sample mean and the null value in units of the sample standard deviation:

  

In other way,

The estimated value of Cohen's d effect size = Mean difference / Sample standard deviation

From the one sample t-test table 2, Mean difference = -3.880

From the one sample t-test table 1, Sample standard deviation = 10.047

Thus Cohen's d effect size

Since cohen's d effect size​ is -0.39 is less than 0.2, this is a small effect.