2 Sample Hypothesis

Question

You wish to test the following claim () at a significance level of .      

      

You obtain 92.6% successes in a sample of size  from the first population. You obtain 96.3% successes in a sample of size  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

The two sample proportions are p1 = 0.926 and p2 = 0.963.

The two sample sizes are n1 = 759 and n2 = 269

First we need to compute the pooled proportion (p) which is the weighted average of the two sample proportions:

(0.,926 * 759 + 0.963 * 269) / (759 + 269) = 0.936

The test statistic is z = (p1 - p2) - 0

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square root(p*(1-p)*(1/n1 + 1/n2))

Since this is a left-tail test the p-value will be equal to P(Z < z).