Question

tn a recent baseball World Series, the Houston Astros were ordered to keep the roof of their stadiun ope . The Houston team claimed that this would make them lose a home-field advantage, becau the noise from fans would be less effective. During the regular season, Houston won 36 of 53 gam played with the roof closed, and they won 15 of 26 games played with the roof open. Treat these results as a simple random sample of games. Use a significance level of 0.05 to test the claim that the proportion of wins at home is higher with a closed roof than with an open roof. Does the closed roof appear to be an advantage? (a) State the null and alternative hypotheses to determine whether the closed roof is an advantage. (b) The test statistic for the above test was z = 0.89 which gives us a p-value of 0.1858. State a conclusion in context of the problem. (c) Based on the results of the hypothesis test, would you expect a confidence interval to include 0. Why or why not?

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P_Closed< P_Open
Alternative hypothesis: P_Closed > P_Open

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.5. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)

p = 0.6456

SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.11453
z = (p₁ - p₂) / SE

b)

z = 0.8934

where p₁ is the sample proportion in sample 1, where p₂ is the sample proportion in sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 0.8934.

Thus, the P-value = 0.1858.

Interpret results. Since the P-value (0.1858) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that proportions of win at home is higher with a closed roof than with an open roof.

c) 95% confidence interval for the difference in proportions is C.I = (- 0.122, 0.3268). Yes from the hypothesis test also we expect the confidence interval to include 0.

$C.I = \left (p_{1}-p_{2} \right )\pm z_{\alpha /2}\times \sqrt{\frac{p_{1}(1-p_{1})}{n_{1}}&plus;\frac{p_{2}(1-p_{2})}{n_{2}}}$

C.I = ( 0.67925 - 0.57692) + 1.96*0.11453

C.I = 0.1023 + 0.2245

C.I = (- 0.122, 0.3268)