Question

You wish to test the following claim (Ha) at a significance level of α=0.10.

      Ho:p1=p2
      Ha:p1<p2

You obtain a sample from the first population with 193 successes and 355 failures. You obtain a sample from the second population with 303 successes and 464 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 critical value for this test? (Report answer accurate to three decimal places.)
critical value =

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

The test statistic is...

• in the critical region
• not in the critical region

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

Answer 1

Hypothesis :

Ho:p1=p2
Ha:p1<p2

a sample from the first population with 193 successes and 355 failures n1 = 193+355 = 548

a sample from the second population with 303 successes and 464 failures n2 = 303 + 464 = 767

p1 = 193 / 548 = 0.3522

q1 = 1-0.3522 = 0.6478

p2 = 303 / 767 = 0.3950

q2 = 1-0.3950 = 0.6050

Mean Diff P1 - P2 = 0.3522 - 0.3950 = - 0.0428

the critical value for this test

Z = 1.2816 at 0.10

Test statistic (two tailed test) :

Z = (P1-P2) / sqrt ( PQ / n1 + PQ / n2)

Here P =(n1p1 + n2p2) / (n1 + n2)

P = ( 548 * 0.3522 + 767 * 0.3950 ) / (548 + 767)

P = 0.3772 and Q = 1 - 0.3772 = 0.6228

PQ = 0.3772 * 0.6228 = 0.2349

Z = -0.0428 / sqrt ( 0.2349/548 + 0.2349/767 )

= -0.0428 / 0.0271

= - 1.5793

Z stat < Z critical

The test statistic is not in the rejection region

The P-Value is .114336. ( two tailed )

The result is not significant at p < .10.

This test statistic leads to a decision to fail to reject the null

There is not sufficient sample evidence to support the claim that the first population proportion is less than the second population proportion