Chapter 9.1, Problem 9.11E

a)

To determine

To find the 95% confidence interval for p

Answer to Problem 9.11E

The 95% confidence interval is, (0.5955,0.8445)

Explanation of Solution

Given:

  $x$ =36

n = 50

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:

Sample proportion,

  $\widehat{p}=\frac{36}{50}=0.72$

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,

  $\begin{array}{l}\left(0.72-1.96\sqrt{\frac{0.72(1-0.72)}{50}},0.72+1.96\sqrt{\frac{0.72(1-0.72)}{50}}\right)\\ \left(0.5955,0.8445\right)\end{array}$

b)

To determine

To explain if we provide more precise interval.

Explanation of Solution

Given:

  $x$ = 36

n = 50

C=confidence level = 0.95

The 95% confidence interval is, (0.5955,0.8445)

The more precise confidence interval contains less possible values of population proportion. Therefore, lower the confidence level, narrower the confidence interval. In this case, we are less confident to sure about the population proportion will fall in that confidence interval. Hence, it can be say that the 90% confidence interval would be more precise than 95%.

c)

To determine

To find the 90% confidence interval for p

Answer to Problem 9.11E

The 90% confidence interval is, (0.6159,0.8241)

Explanation of Solution

Given:

  $x$ = 36

n = 50

C=confidence level = 0.90

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:

Sample proportion,

  $\widehat{p}=\frac{36}{50}=0.72$

Now need to find Zc

The level of significance is = a =0.10

Therefore, Zc = Z _a/2 = 1.96 ..Using excel function, =ABS(NORMSINV(0.10/2))

Therefore, the 90% confidence interval for population proportion is,

  $\begin{array}{l}\left(0.72-1.64\sqrt{\frac{0.72(1-0.72)}{50}},0.72+1.64\sqrt{\frac{0.72(1-0.72)}{50}}\right)\\ \left(0.6159,0.8241\right)\end{array}$

Hence, we are 90% confidence that true population proportion of seniors who attend prom will fall in the above interval.