Question

In the past, 32% of a country club's members brought guests to play golf sometime during the year. Last year, the club initiated a new program designed to encourage members to bring more guests to play golf. In a sample of 90 members, 35 brought guests to play golf after the program was initiated. Therefore, the p-value is .0806. When testing the hypothesis that the new program has increased the proportion of members bringing out guests (using a 10% level of significance), what can you conclude concerning the null hypothesis?

Answer 1

Solution :

Given that ,

n = 90

x = 35

The null and alternative hypothesis is

H0 : p = 0.32

Ha : p > 0.32

This is the right tailed test .

= x / n = 35 / 90 = 0.3889

P0 = 32% = 0.32

1 - P0 = 1 - 0.32 = 0.68

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

= 0.3889 - 0.32 / [0.32 ( 1 - 0.32 ) / 90 ]

= 1.401

The test statistic = 1.401

P-value = 0.0806

= 0.10

0.0806 < 0.10

P-value <

Reject the null hypothesis .

There is sufficient evidence to the test claim .