Question

A study conducted by Harvard Business School recorded the amount of time CEOs devoted to various activities during the workweek. Meetings were the single largest activity averaging 18 hours per week. Assume that the standard deviation for the time spent in meetings is 5.2 hours. To confirm these results, a random sample of 35 CEOs was selected. This sample averaged 16.8 hours per week in meetings. Which of the following statements is correct?
a. The interval that contains 95% of the sample means is 16.3 and 19.7 hours. Because the sample mean is between these two values, we have support for the results of the CEO study by the Harvard Business School.
b. The interval that contains 95% of the sample means is 17.1 and 18.9 hours. Because the sample mean is not between these two values, we do not have support for the results of the CEO study by the Harvard Business School.
c. The interval that contains 95% of the sample means is 15.7 and 20.3 hours. Because the sample mean is between these two values, we have support for the results of the CEO study by the Harvard Business School.
d. The interval that contains 95% of the sample means is 15.7 and 20.3 hours. Because the sample mean is between these two values, we do not have support for the results of the CEO study by the Harvard Business School

Answer 1

Answer:

a. The interval that contains 95% of the sample means is 16.3 and 19.7 hours. Because the sample mean is between these two values, we have support for the results of the CEO study by the Harvard Business School.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 18, \sigma = 5.2, n = 35, s = \frac{5.2}{\sqrt{35}} = 0.879$

95% of the sample means:

From the: 50 - (95/2) = 2.5th percentile.

To the: 50 + (95/2) = 97.5th percentile.

2.5th percentile:

X when Z has a pvalue of 0.025. So X when Z = -1.96.

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$-1.96 = \frac{X - 18}{0.879}$

$X - 18 = -1.96*0.879$

$X = 16.3$

97.5th percentile:

X when Z has a pvalue of 0.975. So X when Z = 1.96.

$Z = \frac{X - \mu}{s}$

$1.96 = \frac{X - 18}{0.879}$

$X - 18 = 1.96*0.879$

$X = 19.7$

95% of the sample means are between 16.3 and 19.7 hours. This interval contains the sample mean of 16.8 hours, which supports the study.

So the correct answer is:

a. The interval that contains 95% of the sample means is 16.3 and 19.7 hours. Because the sample mean is between these two values, we have support for the results of the CEO study by the Harvard Business School.