Question

Suppose that a report indicates that 28% of Canadians have experienced difficulty in making credit card payments. Further suppose that a polling organization randomly sampled 400 Canadians from 10 counties and found that 136 reported such difficulty. Does this indicate that the problem is more severe among these counties? The P-value associated with the test statistic is 0.2119 0.0038 0.025. 0.0177 0.0522.

Answer 1

Given, x=136 n = 400 The sample proportion is, 136 400 =0.34 The null and alternative hypothesis is: Ho: p=0.28 H:p>0.28

Calculate test statistic: Under the null hypothesis, the test statistic for one sample proportion is, -p p(1-P) 0.34-0.28 0.28(1-0.28) V 400 = 2.67 Compute P-value: P-value=P(Z > 2.67) =1-P(Z < 2.67) =1-0.9962 = 0.0038 (Using standard normal table] Correct Answer: Option (b)