In a simple random sample of size 75, there were 19 individuals in the category of interest. It is desired to test $H _ { 0 } : p = 0.38 \text { versus } H _ { 1 } : p

Question

In a simple random sample of size 75, there were 19 individuals in the category of interest. It is desired to test $H _ { 0 } : p = 0.38 \text { versus } H _ { 1 } : p < 0.38$ . Do you reject $H _ { 0 } \text { at the } 0.01$ level?

Quizplus · Accepted Answer

We can use the normal approximation to the binomial distribution since both np (75*0.38) and n(1-p) (75*0.62) are greater than or equal to 10. The test statistic is:

Z = (x - np) / sqrt(np(1-p)/n) = (19 - 75*0.38) / sqrt(75*0.38*0.62/75) = -1.252

Using a one-tailed test with a significance level of 0.01, the critical value is -2.33 (found using a standard normal distribution table or calculator). Since the test statistic (-1.252) is not less than the critical value (-2.33), we fail to reject the null hypothesis. Therefore, we do not have enough evidence to conclude that the true proportion is less than 0.38 at the 0.01 level of significance.

In a Simple Random Sample of Size 75, There Were H0:p=0.38 versus H1:p<0.38H _ { 0 } : p = 0.38 \text { versus } H _ { 1 } : p < 0.38H0​:p=0.38 versus H1​:p<0.38

In a Simple Random Sample of Size 75, There Were $H _ { 0 } : p = 0.38 \text { versus } H _ { 1 } : p < 0.38$