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A shareholders' group is lodging a protest against your company. The shareholders group claimed that the mean tenure for a chief exective office (CEO) was less than 10 years. A survey of 104 companies reported in The Wall Street Journal found a sample mean tenure of 7.7 years for CEOs with a standard deviation of $s=6$ years (The Wall Street Journal, January 2, 2007). You don't know the population standard deviation but can assume it is normally distributed.
You want to formulate and test the hypothesis made by the group, at a significance level of $\alpha=0.002$ . Your hypotheses are:
\[
\begin{array}{l}
H_{o}: \mu \geq 10 \\
H_{a}: \mu< 10
\end{array}
\]
What is the test statistic for this sample?
test statistic $=$
(Repryrt answer accurate to two decimal places.)
What is the $p$-value for this sample?
$\mathrm{p}$-value $=$
(Report answer accurate to four decimal places.)
The p-value is...
less than (or equal to) $\alpha$
greater than $\alpha$
This test statistic leads to a decision to...
reject the null
accept the null
fail to reject the null
As such, the final conclusion is that...
There is sufficient evidence to reject the claim that the population mean is less than 10.
There is not sufficient evidence to reject the claim that the population mean is less than 10 .
There is sufficient evidence to support the claim that the population mean is less than 10.
Answer
As such, the final conclusion is that there is sufficient evidence to support the claim that the population mean is less than 10.
Step 1 ：First, we calculate the test statistic using the formula: \(Z = \frac{M - \mu}{s / \sqrt{n}}\). Here, \(M = 7.7\), \(\mu = 10\), \(s = 6\), and \(n = 104\).
Step 2 ：Substituting the given values into the formula, we get: \(Z = \frac{7.7 - 10}{6 / \sqrt{104}}\).
Step 3 ：Calculating the above expression, we get: \(Z = -5.63\) (rounded to two decimal places). So, the test statistic is \(\boxed{-5.63}\).
Step 4 ：Next, we calculate the p-value. The p-value is the probability that a standard normal random variable is less than -5.63.
Step 5 ：Using a standard normal table or a calculator, we find that the probability that a standard normal random variable is less than -5.63 is approximately 0.0000.
Step 6 ：Since the test is one-tailed, we do not need to multiply this probability by 2 to get the p-value. So, the p-value is \(\boxed{0.0000}\).
Step 7 ：Since the p-value (0.0000) is less than the significance level (0.002), we reject the null hypothesis \(H_{o}\).
Step 8 ：As such, the final conclusion is that there is sufficient evidence to support the claim that the population mean is less than 10.
