Chapter 9.2, Problem 9.36E

a)

To determine

To perform hypothesis test for one sample proportion $\widehat{p}$ .

Answer to Problem 9.36E

There is not sufficient evidence to reject the claim that the coin is fair at 0.05 level of significance.

Explanation of Solution

Given:

Sample size = n = 250

  $x$ = 140

Confidence level = 0.95

Formula:

Sample proportion:

  $\widehat{p}=\frac{x}{n}$

Test statistic:

  $z=\frac{\widehat{p}-p_0}{\sqrt{p_0(1-p_0)/n}}$

Calculation:

The sample proportion is,

  $\widehat{p}=\frac{140}{250}=0.56$

Null and alternative hypothesis:

  $\begin{array}{l}{H}_o:p=0.5\\ H_a:p\ne 0.5\end{array}$

This corresponding to a two-tailed test, for which a z-test for one population proportion needs to be used.

Test statistic:

  $z=\frac{0.56-0.5}{\sqrt{0.5(1-0.5)/250}}=1.90$

Test statistic z = 1.90

Using the P-value approach:

  $2\times P\left(Z<1.90\right)=2\times 0.0287=0.0574$

The p-value is p = 0.0574

Using excel formula, =NORMSDIST(1.90)

And a = 0.05

Since p = 0.0574 >0.05, it is concluded that the Null Hypothesis is not rejected.

Conclusion: There is not sufficient evidence to reject the claim that the coin is fair.

b)

To determine

To find the confidence interval for population proportion.

Answer to Problem 9.36E

The 95% confidence interval is (0.4985, 0.6215)

Explanation of Solution

Given:

Sample proportion: $\widehat{p}=56\%=0.56$

Confidence level = c = 95% = 0.95

Sample size = n = 250

Formula:

Margin of error:

  $E=Z_{\alpha /2}\times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

Confidence interval:

  $(\widehat{p}-E,\widehat{p}+E)$

Confidence level = c = 95% = 0.95

So, level of significance = a = 1-0.95 = 0.05

Z _a/2 = 1.96 …Using excel function, =ABS(NORMSINV(1-0.05/2))

Margin of error is,

  $E=1.96\times \sqrt{\frac{0.56(1-0.56)}{250}}=0.0615$

Therefore, the 95% confidence interval is,

  $(0.56-0.0615,0.56+0.0615)=(0.4985,0.6215)$

Hence, we are 95% confidence that the true population proportion of heads is between 0.4985 and 0.6215.