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, 33 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.)
1. State the distribution to use for the test. (Round your standard deviation to four decimal places.)
P' ~

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

3. What is the p-value? (Round your answer to four decimal places.)

4. Alpha:

5. 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

Answer 1. Since, we are dealing with Proportions, we will use Normal Distribution so,

  

  

Null Hypothesis, Ho: The proportion of a particular state's percent where full-time faculty teaches the online classes is 68% i.e.   

Alternative Hypothesis, Ha: The proportion of a particular state's percent where full-time faculty teaches the online classes is 68% i.e.

Answer 2. Test Statistics is given as:

  

So,

Answer 3. P-value is given as:

P-value = P[ |Z| > 1] = 1 - P[ |Z| <1 ] = 1 - P[ -1 < Z < 1] = 1 - 0.6827 = 0.3173

P- Value = 0.3173

The p-value is greater than the 5% level of significance. Thus we fail to reject the null hypothesis at the this level of significance. We conclude that 68% also represents a particular state's percent for full-time faculty teaching the online classes.

Answer 4. Alpha = 0.05

Answer 5. 95% Confidence Interval for P is given as:

So, 95% Confidence Interval for P is (0.6122,0.8878)