Question

an employment information service claims the mean annual salary for senior level product engineers is $85,000. the annual salaries (in dollars) for a random sample of 16 senior level product engineers are shown in the table to the right. at \LXI0, test the claim that the mean salary is $85,000. complete parts (a) through (e) below. assume the population is normally distributed.

the critical value(s) is/are \\( t_0 = -2.95, 2.95 \\).
(use a comma to separate answers as needed. round to two decimal places as needed.)
choose the graph which shows the rejection region.

(c) find the standardized test statistic \\( t \\).
the standardized test statistic is \\( t = \square \\)
(round to two decimal places as needed.)

Explanation:

Step1: Calculate the sample mean $\bar{x}$

The formula for the sample mean is $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$.
First, sum up all the values in the sample:

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$$

Since $n = 16$, then $\bar{x}=\frac{1507060}{16}=94191.25$.

Step2: Calculate the sample standard deviation $s$

The formula for the sample standard deviation is $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$.

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$$

Then $s=\sqrt{\frac{179977793.5}{16 - 1}}=\sqrt{\frac{179977793.5}{15}}\approx3464.77$.

Step3: Calculate the standardized test statistic $t$

The formula for the $t$-test statistic in a one - sample $t$-test is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$, where $\mu = 95000$, $\bar{x}=94191.25$, $s\approx3464.77$, and $n = 16$.

$$ t=\frac{94191.25-95000}{3464.77/\sqrt{16}}=\frac{- 808.75}{3464.77/4}=\frac{-808.75}{866.19}\approx - 0.93 $$

Answer:

The standardized test statistic $t\approx - 0.93$