Question

You work at a shipyard where employees work two 12-hour shifts each day-shift A and shift B. You have been assigned to study the number of accidents that occur during each shift, to determine if there are any differences. Access the job aid Test Statistic Formulas for formulas you will need as you answer the questions. Select the learning aid Shipyard Accident Cas Study to obtain the data you need, then answer the question. What is the test statistic?

	n	mean	standard deviation
shift A	10	3.7	1.2
shift B	10	3.3	1.4

Answer 1

Solution:-

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

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁ u ₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

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

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 0.5831
DF = 18
t = [ (x₁ - x₂) - d ] / SE

t = 0.686

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 18 degrees of freedom is more extreme than -0.686; that is, less than -0.686 or greater than 0.686.

Thus, the P-value = 0.501

Interpret results. Since the P-value (0.501) is greater than the significance level (0.05), we have to accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is no difference between number of accidents that occur during each shift.