Questions: A shareholders' group is lodging a protest against your company. The shareholders group claimed that the mean tenure for a chief executive office (CEO) was at least 9 years. A survey of 69 companies reported in The Wall Street Journal found a sample mean tenure of 7.4 years for CEOs with a standard deviation of s=5.2 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 a hypothesis that can be used to challenge the validity of the claim made by the group, at a significance level of α=0.002. In other words, your claim is that the mean tenure for a CEO was less than 9 years. Your hypotheses are:
H0: μ ≥ 9
H1: μ<9
What is the test statistic for this sample?
test statistic = -2.5559
What is the p-value for this sample?
p-value = 0.0053
Transcript text: A shareholders' group is lodging a protest against your company. The shareholders group claimed that the mean tenure for a chief executive office (CEO) was at least 9 years. A survey of 69 companies reported in The Wall Street Journal found a sample mean tenure of 7.4 years for CEOs with a standard deviation of $s=5.2$ 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 a hypothesis that can be used to challenge the validity of the claim made by the group, at a significance level of $\alpha=0.002$. In other words, your claim is that the mean tenure for a CEO was less than 9 years. Your hypotheses are:
\[
\begin{array}{l}
H_{o}: \mu \geq 9 \\
H_{1}: \mu<9
\end{array}
\]
What is the test statistic for this sample?
test statistic $=$ $-2.5559$
What is the $p$-value for this sample?
p-value = 0.0053
Solution
Solution Steps
Step 1: Standard Error Calculation
To calculate the standard error \( SE \) of the sample mean, we use the formula:
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{5.2}{\sqrt{69}} \approx 0.626
\]
Step 2: Test Statistic Calculation
The test statistic \( Z_{test} \) is calculated using the formula:
\[
Z_{test} = \frac{\bar{x} - \mu_0}{SE} = \frac{7.4 - 9}{0.626} \approx -2.5559
\]
Step 3: P-value Calculation
For a left-tailed test, the p-value is determined using the standard normal distribution:
\[
P = T(z) \approx 0.0053
\]
Final Answer
The test statistic is \( Z_{test} = -2.5559 \) and the p-value is \( P \approx 0.0053 \).
Thus, the final answers are:
\[
\boxed{Z_{test} = -2.5559}
\]
\[
\boxed{P = 0.0053}
\]