Question

You wish to test the following claim ( H a ) at a significance level of α = 0.005

H o : p 1 = p 2

H a : p 1 < p 2

You obtain 91.1% successes in a sample of size n 1 = 651 from the first population. You obtain 94.8% successes in a sample of size n 2 = 621 from the second population. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution.

What is the test statistic for this sample? (Report answer accurate to three decimal places.)

test statistic =

What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value =

Answer 1

Solution :

The null and alternative hypotheses are as follows :

To test the hypothesis we shall use z-test. The test statistic is given as follows :

Sample proportion of first sample is,

Sample proportion of second sample is,

n​​​​​​1 = 651

n​​​​​​2 = 621

Q = 1 - 0.929 = 0.071

The value of the test statistic is -2.569.

Since, our test is left-tailed test, therefore we shall obtain left-tailed p-value for the test statistic. The p-value is given as follows :

p-value = P(Z < value of the test statistic)

p-value = P(Z < -2.569)

p-value = 0.0051

The p-value is 0.0051.

Significance level (α) = 0.005

Since, p-value is greater than the significance level of 0.005, therefore we shall be fail to reject the null hypothesis (H​​​​​​0) at 0.005 significance level.

Conclude : There is not sufficient evidence to support the claim that p1 < p2.