Question

Given the null and alternative hypotheses below, test the hypothesis using a = 0.01 assuming that a sample of n = 200 yielded x = 106 items with the desired attribute. Ho p20.59 HA P<0.59 Check the requirement for a hypothesis test for a proportion In this situation, np is (Type integers or decimals.) 5 and n(1-P)= 75. Thus, the requirement satisfied What is the test statistic? z= (Round to two decimal places as needed) What is the p-value? p-value = (Round to three decimal places as needed) What is the proper conclusion? Ho. There is evidence to conclude that the population proportion is 059

Answer 1

Now,

$np=200(0.59)=118>5$ and $n(1-p)=200(1-0.59)=82>5$

So, np = 118 is greater than 5 and n(1-p) = 82 is greater than 5. Thus, requirement for normal approximation is satisfied.

The test statistic value is given by:

$Z=\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{\frac{106}{200}-0.59}{\sqrt{\frac{0.59(1-0.59)}{200}}}=-1.73$

Since this is a left tailed test, so the p-value is given by:

$P(Z<-1.73)=P(Z>1.73)=1-P(Z<1.73)=1-0.95818$

$=0.04182\approx 0.042$

Since p-value is greater than 0.01, so we do not have sufficient evidence to reject the null hypothesis H0.

Thus, we do not reject H0. There isn't sufficient evidence to conclude that the population proportion is less than 0.59.