Question

is there a doctor in the house? a market research firm reported the mean annual earnings of all family practitioners in the united states was $178,258. a random sample of 44 family practitioners in los angeles had mean earnings of \LXI0 with a standard deviation of $41,407. do the data provide sufficient evidence to conclude that the mean salary for family practitioners in los angeles differs from the national average? use the \\( \alpha = 0.01 \\) level of significance and the p - value method and excel.\
part: 0 / 5\
part 1 of 5\
(a) state the appropriate null and alternate hypotheses.\
\\( h_0: \mu = 178258 \\)\
\\( h_1: \mu \
eq 178258 \\)\
this hypothesis test is a two - tailed test.\
part: 1 / 5\
part 2 of 5\
(b) compute the value of the test statistic. round the answer to at least three decimal places.\
\\( t = \square \\)

Explanation:

Step1: Recall t - test formula

The formula for the t - test statistic in a one - sample t - test is \(t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size.

From the problem:

• \(\bar{x} = 193250\)
• \(\mu=178258\) (from the null hypothesis \(H_0:\mu = 178258\))
• \(s = 41407\)
• \(n = 44\)

Step2: Calculate the numerator

First, calculate \(\bar{x}-\mu\):
\(\bar{x}-\mu=193250 - 178258=14992\)

Step3: Calculate the denominator

Calculate \(\frac{s}{\sqrt{n}}\):
\(\sqrt{n}=\sqrt{44}\approx6.6332\)
\(\frac{s}{\sqrt{n}}=\frac{41407}{6.6332}\approx6242.35\)

Step4: Calculate the t - statistic

Now, calculate \(t\) using the formula \(t = \frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\):
\(t=\frac{14992}{6242.35}\approx2.402\) (rounded to three decimal places)

Answer:

\(t\approx2.402\)