Chapter 9.1, Problem 9.12E

a)

To determine

To find the 95% confidence interval for p

Answer to Problem 9.12E

The 95% confidence interval is, (0.1446,0.2554)

Explanation of Solution

Given:

n = 200

C=confidence level = 0.95

Formula:

Sample proportion:

  $\widehat{p}=\frac{x}{n}$

Confidence interval for population proportion is,

  $CI=\widehat{p}\mp Zc\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

Calculation:

Here, $x$ is the number of times a 0 occurs at the first digit of a five-digit group. So,

  $x$ = 36

Sample proportion,

  $\widehat{p}=\frac{40}{200}=0.2$

Now need to find Zc

The level of significance is = a =0.05

Therefore, Zc = Z _a/2 = 1.96 ..Using excel function, =ABS(NORMSINV(0.05/2))

Therefore, the 95% confidence interval for population proportion is,

  $\begin{array}{l}\left(0.2-1.96\sqrt{\frac{0.2(1-0.2)}{200}},0.2+1.96\sqrt{\frac{0.2(1-0.2)}{200}}\right)\\ \left(0.1446,0.2554\right)\end{array}$

b)

To determine

To explain if confidence interval included p = 0.1

Explanation of Solution

Given:

The 95% confidence interval is, (0.1446,0.2554)

The confidence interval (0.1446,0.2554) does not contain p = 0.1. Hence, it would be indicated that the first digit in each of the 200 five-digit groups of table B were not randomly selected.