A market research firm reported that the mean annual earnings of all family practitioners in the United States was $179,574. A random sample of 38 family practitioners in New York that month Had mean earnings of $\bar { x } = \$ 198,513$ with a standard deviation of $35,113. You wish to test whether Family practitioners in New York make more than the national average. State a conclusion regarding H₀. Use the $\alpha = 0.1$ level of significance.

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We need to conduct a one-sample t-test with null hypothesis H₀: $\mu \leq \$179,574$ and alternative hypothesis HA: $\mu > \$179,574$, where $\mu$ is the population mean earnings of family practitioners in New York. We can use a t-distribution since the sample size is less than 30 and the population standard deviation is unknown. The test statistic is given by: $t = \frac{\bar{x} - \mu_{0}}{s/\sqrt{n}} = \frac{(\$198,513 - \$179,574)}{\$35,113 / \sqrt{38}} = 2.35$ Using a t-table or calculator with 37 degrees of freedom (df = n-1), the critical t-value for a one-tailed test with $\alpha = 0.1$ is 1.314. Since the calculated t-value is greater than the critical t-value (2.35 > 1.314), we reject the null hypothesis and conclude that there is evidence to suggest that family practitioners in New York make more than the national average. Therefore, the answer is B.

A Market Research Firm Reported That the Mean Annual Earnings xˉ=$198,513\bar { x } = \$ 198,513xˉ=$198,513

A Market Research Firm Reported That the Mean Annual Earnings $\bar { x } = \$ 198,513$