Chapter PVII, Problem 6RE

(a)

To determine

To construct: the 95 percent confidence interval for true distance.

Answer to Problem 6RE

(36.21, 60.51)

Explanation of Solution

Given:

   Formula used:

  $C.I.=\overline{x}+t\times \frac{s}{\sqrt{n}}$

Calculation:

  

(b)

To determine

To explain: the mean distance is plausibly 40 feet, yes or no on the basis of confidence interval.

Explanation of Solution

40 ft is plausible since 40 value comes in the 95 percent confidence interval

= (36.2142, 60.5130).

(c)

To determine

To find: the 99 percent confidence interval for true distance differ form the actual estimating in part (a) and discuss the brief, without actual estimating a new interval.

Explanation of Solution

The 99 percent confidence interval is longer than the confidence interval of 95 percent. This is because the 99 percent reliability factor t value is higher than the confidence level of 95 percent.

(d)

To determine

To explain: at the same confidence interval, how big a sample size does the student want to get a confidence interval half as broad as the one you received in part (a).

Explanation of Solution

To raise the sample size by 4 times, $4\times 11=44$ , to get the confidence interval half.Narrower intervals are created by larger sample sizes. There is an inverse square root association between confidence intervals and sample sizes.