Question

Leisure Time: In a Gallup poll, 1200 adults were randomly selected and asked if they were satisfied or dissatisfied with the amount of leisure time that they had. Of this Sample 690 said that they were satisfied and 510 said they were dissatisfied. Use a 0.01 significance level to test the claim that less than 60% of adults are satisfied with the amount of leisure time that they have.

What is the p-value and the test statistic?

Do we reject the null or fail to reject the null?

Write a conclusion.

Answer #1

Ho: P = 0.6

Ha: P ≤ 0.6

given xo = 690/1200 = 0.575, so 1-xo = 0.425

In such a binary distribution, variance = xo × (1-xo) / n = 0.575 × 0.425 / 1200 = 2.0365 × 10-4

sd or σ = 0.0143

t-value = (xo − M) / σ/(√ n) = |(0.575 - 0.6)| / 0.0143/(√1200) = 60.69

p value is 0 for this case from the table or using any software

A small **p**-value (in our case ≤ 0.01) indicates
strong evidence against the null hypothesis, so I am rejecting the
null hypothesis. Since, we also observe that 0.575 < 0.6 we can
confirm the one-tailed scenario and confirm that we fail to reject
the claim that less than 60% of adults are satisfied with the
amount of leisure time that they have.

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