A recent report estimated that 25% of all college students in the United States have a sexually transmitted disease (STD). Due to the demographics of the community, the director of the campus health center believes that the proportion of students who have a STD is higher at his college.
He tests H0: p = 0.25 versus Ha: p > 0.25. The campus health center staff select a random sample of 150 students and determine that 43 have been diagnosed with a STD.
Conduct a hypothesis test to address the director’s hypothesis. Use a 5% significance level to make your decision. Use the applet to determine the P?value.
Which of the following is an appropriate conclusion based on the results?
A. The data provides strong evidence that the proportion of students at his college who have a STD is more than 25%.
B. The data suggests that the proportion of students at his college who have a STD is 29%.
C. Of the students surveyed at his college, 29% have an STD, but this is not strong enough evidence to conclude that the proportion of students at his college who have a STD is higher than the report’s estimate.
D.This survey suggests that 25% of students at his college have a STD.

Decision: we FAIL to REJECT the null hypothesis. That is, there is no convincing evidence that the proportion of students at his college who have a STD is more than 25%. Therefore the null hypothesis is valid.

Rule

If;

P-value > significance level --- accept Null hypothesis

P-value < significance level --- reject Null hypothesis

Z score > Z(at 95% confidence interval) ---- reject Null hypothesis

Z score < Z(at 95% confidence interval) ------ accept Null hypothesis

Step-by-step explanation:

Given;

n=150 represent the random sample taken

Null hypothesis: H0 = 0.25

Alternative hypothesis: Ha > 0.25

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size = 150

po = Null hypothesized value = 0.25

p^ = Observed proportion = 43/150 = 0.287

Substituting the values we have

z = (0.287-0.25)/√{0.25(1-0.25)/150}

z = 1.046518

z = 1.047

To determine the p value (test statistic) at 0.05 significance level, using a one tailed hypothesis.

P value = P(Z<1.047) = 0.14755 = 0.148

Since z at 0.05 significance level is between -1.96 and +1.96 and the z score for the test (z = 1.047) which falls within the region bounded by Z at 0.05 significance level. And also the one-tailed hypothesis P-value is 0.148 which is higher than 0.05. Then we can conclude that we don't have enough evidence to REJECT the null hypothesis, and we can say that at 5% significance level the null hypothesis is valid.