Question

When games were sampled throughout a​ season, it was found that the home team won 116 of 153 baseball ​games, and the home team won 59 of 66 hockey games. The result from testing the claim of equal proportions are shown on the right. Does there appear to be a significant difference between the proportions of home​ wins? What do you conclude about the home field​ advantage? ​2-proportion test p 1 not equals p 2 z equals negative 2.30089323 p dash value equals 0.02139767 ModifyingAbove p with caret 1 equals 0.7581699346 ModifyingAbove p with caret 2 equals 0.8939393939 p overbar equals 0.7990867580

Does there appear to be a significant difference between the proportions of home​ wins? (Use the level of significance alpha equals 0.05 . right parenthesis A. Since the​ p-value is large​, there is not a significant difference. B. Since the​ p-value is large​, there is a significant difference.

Answer 1

Answer)

Ho : P1 = P2

Ha : P1 is not equal to P2

N1 = 153, P1 = 116/153

N2 = 66, P2 = 59/66

First we need to check the conditions of normality that is if n1p1 and n1*(1-p1) and n2*p2 and n2*(1-p2) all are greater than equal to 5 or not

N1*p1 = 116

N1*(1-p1) = 37

N2*p2 = 59

N2*(1-p2) = 7

All the conditions are met so we can use standard normal z table to conduct the test

Test statistics z = (P1-P2)/standard error

Standard error = √{p*(1-p)}*√{(1/n1)+(1/n2)}

P = pooled proportion = [(p1*n1)+(p2*n2)]/[n1+n2]

After substitution

Test statistics z = -2.3

From z table, P(Z<-2.3) = 0.0107

So, P-Value = 0.0107

As the obtained P-Value is less than the given significance (0.05)

We reject the null hypothesis

So, we can conclude that there is a difference