Chapter 8, Problem 12PT

a.

To determine

To check: If the conditions of normality are satisfied or not.

Answer to Problem 12PT

Yes.

Explanation of Solution

Given:

Sample size = 48

Sample mean = 6.208

Sample standard deviation = 2.576

Formula used:

The formula to compute the confidence interval is:

  $\overline{x}-t_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{n}}<\mu <\overline{x}+t_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{n}}$

To construct the confidence interval either the sample size should be 30 or more or the population must be distributed normal.

Here, sample size is above 30. Thus, it could be concluded that the normality assumption is met.

b.

To determine

To find: The mistake in the confidence interval of Maurice and compute the 95% confidence interval.

Answer to Problem 12PT

The confidence interval is (5.4565, 6.9594).

Explanation of Solution

Calculation:

Maurice has computed z-interval instead of t-interval and also, he has taken incorrect standard deviation.

The degree of freedom is 48-1 = 47

From the t-table, the critical value at 95% confidence level is 2.021.

The 95% confidence interval could be computed as:

  $\begin{array}{c}6.208-2.021\times \frac{2.576}{\sqrt{48}}<\mu <6.208+2.021\times \frac{2.576}{\sqrt{48}}\\ 5.4565<\mu <6.9594\end{array}$

Thus, the confidence interval is (5.4565, 6.9594).