Question

Student Life: Employment Professor Jennings claims that only 35% of the students at Flora College work while attending school. Dean Renata thinks that the professor has underestimated the number of students with part-time or full time jobs. A random of 81 students shows that 39 have jobs. Do the data indicate that more than 35% of the students have jobs? (Use a 5% level of significance.)

1. What is the level of significance? State the null hypothesis and alternate hypotheses.
2. Check Requirements What sampling distribution will you use? What assumptions are you making? What is the value of the sample test statistic?
3. Find (or estimate) the P-value. Sketch the sampling distribution and show the area corresponding to the P-value
4. Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level a?
5. Interpret your conclusion in the context of the application.

Note: For degrees of freedom d.f. not in the Student’s t table, use the closing d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more conservative answer. Answers may vary due to rounding.

Answer 1

Solution:

Given: Employment Professor Jennings claims that only 35% of the students at Flora College work while attending school.

Sample size = n = 81

x = Number of students have jobs = 39

Thus sample proportion is:

We have to test if  the data indicate that more than 35% of the students have jobs or not.

Level of significance =

Part a) What is the level of significance? State the null hypothesis and alternate hypotheses.

Level of significance =

Since we have to test more than 35% of the students have jobs, this is right tailed test.

Part b) Check Requirements:

What sampling distribution will you use? What assumptions are you making?

Since Sample size = n = 81 is large and

n * p = 81 * 0.35 = 28.35 > 10 and n * ( 1 - p) = 81 * ( 1 - 0.35) = 81 * 0.65 = 52.65 > 10

both the assumpions are satisfied and sample is randomly selected. Thus we use Normal approximation to proporion.

What is the value of the sample test statistic?

Part c) Find (or estimate) the P-value. Sketch the sampling distribution and show the area corresponding to the P-value

P-value = P( Z > z test statistic value)

P-value = P( Z > 2.48 )

P-value = 1 - P( Z < 2.48 )

Look in z table for z = 2.4 and 0.08 and find area.

P( Z < 2.48 ) = 0.9934

thus

P-value = 1 - P( Z < 2.48 )

P-value = 1 - 0.9934

P-value = 0.0066

Part d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level a?

Decision Rule: Reject H0, if P-value < 0.05 level of significance , otherwise we fail to reject H0.

Since P-value = 0.0066 < 0.05 level of significance, we reject null hypothesis H0. Thus data is statistically significant at level 0.05.

Part e) Interpret your conclusion in the context of the application.

Since we have rejected null hypothesis H0, we conclude that: the data indicate that more than 35% of the students have jobs.

Thus that Professor Jennings has underestimated the number of students with part-time or full time jobs.