Question

Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data.

Short answer

Confidence Interval: (0.65, 0.90)

Step-by-step solution

Find H0 and Ha: We want to test if California community colleges take 68% of the online classes thought by full-time faculties, or not.

$H_0:p=0.68;H_a:p\ne 0.68$

Determine the distribution needed: In words, CLEARLY state what your random variable P′ represents.Let P′ = the proportion of online courses at LBCC that are taught by full-time faculty.

Normal:$N\left(0.68,\sqrt{\frac{(0.68)(1-0.68)}{44}}\right)$

Test Statistic:$z=0.1873$

Calculate the p-value using the normal distribution for proportions:

$p-value=0.1873$

In one to two complete sentences, explain what the p-value means for this problem.

If the null hypothesis is true (the proportion is 0.68), then there is a 0.1873 probability that the sample (estimated) proportion is 0.773 or more.

Compare α and the p-value:Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences.

alpha	decision	reason for decision
0.05	Do not reject the null hypothesis.	$p-value>0.05$

Conclusion: At the 5% level, the data do not provide statistically significant evidence that the true proportion of online courses taught by full-time faculty is not 68%.

Confidence Interval

Confidence Interval: (0.65, 0.90): The “plus-4s” confidence interval is (0.6275, 0.8725).