Question

1. (5pt) In a recent New York City marathon, 25,221 men finished and 253 dropped out. Also,

12,883 women finished and 163 dropped out. We want to use a 0.01 significance level to test

the claim that the rate of those who finish is the same for men and women.

a) Test the claim using a hypothesis test (identify the null hypothesis, alternative hypothesis,

test statistic, critical value (or P-value) and conclusion)

b) Test the claim by constructing an appropriate confidence interval.

Answer 1

Answer:

(a) Explained below.

(b) The 99% confidence interval for the difference between proportions is (-0.00094, 0.00494).

Step-by-step explanation:

The information provided is:

n (Men who finished the marathon) = 25,221

n (Women who finished the marathon) = 12,883

Compute the proportion of men and women who finished the marathon as follows:

$\hat p_{1}=\frac{25221}{25221 +253}=0.99\\\\\hat p_{2}=\frac{12883 }{12883+163}=0.988$

The combined proportion is:

$\hat P=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}\\\\=\frac{25221+12883}{38520}\\\\=0.989$

(a)

The hypothesis is:

H₀: The rate of those who finish the marathon is the same for men and women, i.e. p₁ - p₂ = 0.

Hₐ: The rate of those who finish the marathon is not same for men and women, i.e. p₁ - p₂ ≠ 0.

Compute the test statistic as follows:

$Z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat P(1-\hat P)\cdot [\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}$

   $=\frac{0.99-0.988}{\sqrt{0.989(1-0.989)\times[\frac{1}{25474}+\frac{1}{13046}]}}\\\\=1.78$

Compute the p-value as follows:

$p-value=2\times P(Z>1.78)\\\\=2\times 0.03754\\\\=0.07508\\\\\approx 0.075$

The p-value of the test is more than the significance level. The null hypothesis was failed to be rejected.

Thus, concluding that the rate of those who finish is the same for men and women.

(b)

Compute the 99% confidence interval for the difference between proportions as follows:

The critical value of z for 99% confidence level is 2.58.

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\cdot\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

     $=(0.99-0.988)\pm 2.58\times\sqrt{\frac{0.99(1-0.99)}{25474}+\frac{0.988(1-0.988)}{13046}}\\\\=0.002\pm 0.00294\\\\=(-0.00094, 0.00494)$

Thu, the 99% confidence interval for the difference between proportions is (-0.00094, 0.00494).