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?
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 .
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