Question

You wish to test the following claim ( H a ) at a significance level of α = 0.10 . H o : p 1 = p 2 H a : p 1 > p 2

You obtain a sample from the first population with 133 successes and 545 failures. You obtain a sample from the second population with 22 successes and 202 failures.

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 =

The p-value is...

• less than (or equal to) αα
• greater than αα

This test statistic leads to a decision to...

• reject the null
• accept the null
• fail to reject the null

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the claim that the first population proportion is greater than the second population proportion.
• There is not sufficient evidence to warrant rejection of the claim that the first population proportion is greater than the second population proportion.
• The sample data support the claim that the first population proportion is greater than the second population proportion.
• There is not sufficient sample evidence to support the claim that the first population proportion is greater than the second population proportion.

What function do I use in my calculator for this because it is a double sample? Thank you!

Answer 1

α = 0.10

n1 = 133+545 = 678

n2 = 22+202 = 224

Given =133 /678 = 0.196

= 22/224 = 0.098

Let p1 and p2 be the population proportions.

We have to check if p1 > p2.

This corresponds to a right-tailed test, for which a z-test for two population proportions needs to be conducted.

So accordingly the null and the alternate hypothesis would be:

pooled proportion = (133+22)/678+224 = 0.172

Standard error (SE) = = = 0.0289

Z-statistic would be = (0.196 - 0.098)/ 0.0289 = 3.39

For z = 3.39, p-value comes out to be 0.0003 [using invNorm function in calculator]

Since p < α = 0.10

Hence we have strong evidence against the null hypothesis. Hence the null hypothesis would be rejected.

Conclusion:The sample data support the claim that the first population proportion is greater than the second population proportion.