** ONE SAMPLE T-TEST**

One sample t-test is used to determine if the mean of a sample is different from a particular value for a single population. Continue with the One Sample T-Test Case Analysis.

CASE ANALYSIS-1

** ****PROBLEM**

A consultancy firm claims that the average salary offered by its firm to the MBA fresher’s is Rs. 22,000 per month. A sample of 15 management students of a college was taken and the information was collected about their starting salary, to test the claim of the firm.

**Table-1:** Sample Data

The hypotheses for the analysis are:

Null hypothesis-H_{0}: The average salary of MBA fresher’s is Rs.22, 000. (μ= 22,000)

Alternative Hypothesis- H_{1}: The average salary of MBA fresher’s is not Rs.22, 000 (μ 22,000) (two-tailed test)

**Input Data**

The salaries of 15 management students are treated as input data for the analysis.

**Performing the Analysis with SPSS**

For SPSS Version 11, click on **Analyze ⇒ ****Compare Means ⇒ One-Sample T test. **This will bring up the SPSS screen dialogue box as shown below.

After clicking **One Sample T test, **this will bring up the following SPSS screen dialogue box.

Select the dependent variable and click it to move it to **Test variables** box. In this case, we are comparing if the average salary is Rs.22, 000 or not. So we should enter 22,000 into the Test Value box.

Finally click OK of the main dialogue box.

** ****SPSS Output**

The SPSS outputs of the analysis are given in following tables.

The table tells that the average salary of 15 MBA fresher’s is Rs19, 600 with standard deviation 4595.02837.

**From the output, ****t ****= -2.203 with 14 degrees of freedom.**

**DECISION**

Reject the null hypothesis if p-value (Sig. (2-tailed)) ≤ 0.05

**INTERPRETATION**

The p-value is 0.063 and it is more than 0.05 (**5% level of significance**), so we accept the null hypothesis and reject the alternative hypothesis at 5% level of significance. It is concluded that the average salary of MBA fresher’s offered by the consultancy firm is not Rs. 22,000.

**CASE ANALYSIS-2**

** ****PROBLEM**

The average sale of a chocolate bar of brand X is 130 dozens per month. After an advertisement campaign the sale is claimed to increase. A sample of 20 stores in a locality is chosen to test the validity of the claim. The sale of chocolate bars per store is given below.

**Table-1:** Sample Data

Null hypothesis-H_{0}: The average sale of chocolate bars per store is 130 dozens (μ 130 dozens).

Alternative Hypothesis- H_{1}: The sale average sale of chocolate bars per store is more than 130 dozens (μ 130 dozens). **(One-tailed test)**

** ****SPSS Output**

**From the output, ****t ****= -.136 with 21 degrees of freedom.**

**DECISION**

Reject the null hypothesis if p-value (Sig. (2-tailed)) ≤ 0.05

The p-value for one-tailed test would be (Sig. (2-tailed))/2

**INTERPRETATION**

The p-value is (Sig. (2-tailed))/2 = 0.893/2 = 0.446 and it is less than 0.05 (**5% level of significance**), therefore we reject the null hypothesis and accept the alternative hypothesis at 5% level of significance. Therefore it is concluded that the mean sales of chocolate bar has increased due to the advertisement campaign.

**SPSS Command**

- Click on ANALYZE at the SPSS menu bar (in older versions of SPSS, click on STATISTICS instead of ANALYZE).
- Click on COMPARE MEANS followed by ONE SAMPLE T TEST.
- Select the test variable and move it TEST VARIABLE(S) box. Fill the box TEST VALUE with appropriate hypothesized value and click CONTINUE.
- Select OK of the main dialogue box.