A company claims that the average salary of its employees is $75,000. A random sample of 100 employees is taken and their average salary is found to be $72,000 with a standard deviation of $5000. Test the company's claim using a significance level of 0.05.
Solution
Step 1 ：Step 1: State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis. \n Null hypothesis: \(\mu = 75000\) \n Alternative hypothesis: \(\mu \neq 75000\)
Step 2 ：Step 2: Formulate an analysis plan. For this analysis, the significance level is defined as 0.05. The test statistic is a z-score (z).
Step 3 ：Step 3: Calculate the test statistic. \n \(z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{72000 - 75000}{\frac{5000}{\sqrt{100}}} = -6\)
Step 4 ：Step 4: Interpret the results. If the test statistic is outside of the critical region, we accept the null hypothesis. Otherwise, we reject it. Since our calculated z-score (-6) is less than the critical value for a two-tailed test at the 0.05 level (approximately ±1.96), we must reject the null hypothesis.
