T-Test Report

T-Test Report

The production satisfaction requirements have been set at a 95% confidence level that the target market will have a product satisfaction score of 4.5. In this observation study using a simple random sample of people; the people were asked to take the product home for one month and rate their satisfaction on a scale of 1 to 5 with 1 being unsatisfactory and 5 being very satisfied. Recognizing the value of setting the satisfaction confidence level so high, the team has used the null and alternative hypothesis using the following formulas and questions.

H0 : To arrive at the null hypothesis the average satisfaction must be > 4.5 in other words the Zippy Dippy Can Opener performs exactly how they expected it to perform.

H1: To arrive at the alternative hypothesis the average satisfaction must be < 4.5 in other words the Zippy Dippy Can Opener does not perform as expected or promised.

The test we performed is a 1-tailed test. Due to the fact we were asked to prove a satisfaction level of 4.5 we are interested in the fact that the satisfaction score is above and below. It is irrelevant if the score is 4.5, 4.6, or higher it just needs to be above 4.5. A 2-tailed test would have us examining the numbers on either end of the 2 satisfaction extremes. Understanding the formulas of the t-statistic we look at the following which includes the satisfaction score of 4.5 at a 95% satisfaction level with a sample size of 20, an average score of 4.7, and a standard deviation of 0.25:

t = [x- u] / (s/n1/2) t= (4.7 - 4.5) = 3.578

0.25

4.47

The next step is to look up the t-statistic from the t-table. To use the t-distribution table we determine the degrees of freedom in the sample size minus 1 or 20-1=19. From the t-table in the one tail for an area of 0.025 we find the value of the t-statistic to be 2.093. We compare the t-statistic calculated using the test data or 3.578 with the t-statistic determined by using the t-distribution table at 19 degrees of freedom and a 95% confidence level or 2.093. Since the 3.578 is to the right of 2.093 we accept the null hypothesis. Thus we can say with a 95% confidence that the customer satisfaction score will be at least 4.5.

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