Question

In a particular year, 68% of online courses taught at a system of community colleges were taught by full-time faculty. To test if 68% also represents a particular state's percent for full-time faculty teaching the online classes, a particular community college from that state was randomly selected for comparison. In that same year, 31 of the 44 online courses at this particular community college were taught by full-time faculty. Conduct a hypothesis test at the 5% level to determine if 68% represents the state in question.

Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

(A). In words, state what your random variable P' represents.

(B). State the distribution to use for the test. (Round your standard deviation to four decimal places.)

P' ~ ____ ( ____ , ____ ).

(C). What is the test statistic? (Round your answer to two decimal places.)

t = _____.

(D). What is the p-value? (Round your answer to four decimal places.) Explain what the p-value means for this problem.

(E). Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value.

(F). Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion.

(i) Alpha:

(ii) Decision:

(iii) Reason for decision:

(iv) Conclusion:

(G) Construct a 95% confidence interval for the true proportion. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your answers to four decimal places.)

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P = 0.68
Alternative hypothesis: P 0.68

Note that these hypotheses constitute a two-tailed test.  

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).

S.D = sqrt[ P * ( 1 - P ) / n ]

S.D = 0.07032
z = (p - P) /S.D

z = 0.35

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -0.35 or greater than 0.35.

Thus, the P-value = 0.726.

Interpret results. Since the P-value (0.726) is greater than the significance level (0.05), we have to acept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that 68% represents the state in question.