Question

In the past, 30% 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 80 members, 29 brought guests to play golf after the program was initiated. Therefore, the test statistic is 1.22. When testing the hypothesis that the new program has increased the proportion of members bringing out guests (using a 5% level of significance), what is the p-value? (please round your answer to 4 decimal places)

Answer 1

Solution:

Given: In the past, 30% of a country club's members brought guests to play golf sometime during the year.

That is: p = 0.30

Sample Size = n = 80

x = Number of guests brought to play golf after the program was initiated = 29

We are also given that test statistic value = z = 1.22

We have to test the hypothesis that the new program has increased the proportion of members bringing out guests.

That is we have to test if population proportion p is greater than 0.30

Thus hypothesis of the study are:

H₀: p = 0.30 Vs H₁: p > 0.30

We have to find p-value for this hypothesis test.

Since H₁: p > 0.30 , is right tailed, p-value is given by:

p-value = P( Z > z test statistic value)

p-value = P( Z > 1.22 )

p-value = 1 - P( Z < 1.22 )

Look in z table for z = 1.2 and 0.02 and find corresponding area.

P( Z < 1.22 ) = 0.8888

Thus we get:

p-value = 1 - P( Z < 1.22 )

p-value = 1 - 0.8888

p-value = 0.1112

Decision rule: Reject H₀, if p-value < 0.05 level of significance, otherwise we fail to reject H₀.

Since p-value = 0.1112 >   0.05 level of significance, we fail to reject H₀.

Conclusion: Since we failed to reject H₀, there is not sufficient evidence to conclude that:  the new program has increased the proportion of members bringing out guests.