A market research firm reported that the mean annual earnings of all family practitioners in the United States was $178,695. A random sample of 53 family practitioners in New York that month Had mean earnings of $\bar { x } = \$ 197,064$with a standard deviation of $35,332. You wish to test whether Family practitioners in New York make more than the national average.
Compute the value of the test statistic and state the number of degrees of freedom.

Question

Quizplus · Accepted Answer

The test statistic for comparing a sample mean to a population mean is calculated using the formula $t = \frac{\bar{x} - \mu}{(s/\sqrt{n})}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $s$ is the sample standard deviation, and $n$ is the sample size. Here, $\bar{x} = \$197,064$, $\mu = \$178,695$, $s = \$35,332$, and $n = 53$. Plugging these values into the formula gives a test statistic of 3.785. The degrees of freedom for a t-test is $n - 1$, which is $53 - 1 = 52$.

A Market Research Firm Reported That the Mean Annual Earnings xˉ=$197,064\bar { x } = \$ 197,064xˉ=$197,064

A Market Research Firm Reported That the Mean Annual Earnings $\bar { x } = \$ 197,064$