Question

a) 25 CEOs in the healthcare industry are polled. The average salary with stock bonuses for the poll is $282,000 with a sample std. dev. of $71,000. What is the 95% confidence interval for the population mean of CEOs salary with stock bonuses for the healthcare industry?

b) 25 CEOs in the healthcare industry are polled. The average salary with stock bonuses for the poll is $282,000 with a sample std. dev. of $71,000. It is known that in 1990, the average salary was $250,500 (adjusted for inflation). Is there evidence that the salary for CEOs in healthcare have increased?

Answer 1

a)

We need to construct the 95% confidence interval for the population mean μ. The following information is provided 282000 71000 25 Sample Mean X = Sample Standard Deviation (s)- Sample Size (n) - The critical value for α 0.05 and df -1 24 degrees of freedom is tc-21-a/21-1 The corresponding confidence interval is computed as shown below: 2.064 71000 71000 25 = 282000-2.064 -- , 282000 2.064 25 (252692.64, 311307.36) Therefore. based on the data provided, the 95% confidence interval for the population mean is 252692.64 < < 311307.36, which indicates that we are 95% confident that the true population mean u is contained by the interval (252692.64,311307.36)

b) Assuming level of significance = 0.05

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 250500

Ha: μ > 250500

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is tc​=1.711.

The rejection region for this right-tailed test is R=t:t>1.711

(3) Test Statistics

The t-statistic is computed as follows:

$t = \frac{\bar X - \mu_0}{s/\sqrt{n}} = \frac{ 282000 - 250500}{ 71000/\sqrt{ 25}} = 2.218$

(4) Decision about the null hypothesis

Since it is observed that t=2.218>tc​=1.711, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0181, and since p=0.0181<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is greater than 250500, at the 0.05 significance level.

Graphically

T-Test Results:tstats-2.218,p-value -0.0181 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 4.0 -3.5-3.0 -2.5 -2.0 15 1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

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