Question

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 sample of 85 students shows that 40 have jobs. Do the data indicate that more than 35% of the students have jobs? Use a 5% level of significance.

What are we testing in this problem?

single meansingle proportion    

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: p = 0.35; H₁: p ≠ 0.35H₀: μ = 0.35; H₁: μ > 0.35    H₀: μ = 0.35; H₁: μ < 0.35H₀: p = 0.35; H₁: p < 0.35H₀: p = 0.35; H₁: p > 0.35H₀: μ = 0.35; H₁: μ ≠ 0.35

(b) What sampling distribution will you use? What assumptions are you making?

The Student's t, since np > 5 and nq > 5.The standard normal, since np > 5 and nq > 5.    The Student's t, since np < 5 and nq < 5.The standard normal, since np < 5 and nq < 5.

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


(c) Find (or estimate) the P-value.

P-value > 0.2500.125 < P-value < 0.250    0.050 < P-value < 0.1250.025 < P-value < 0.0500.005 < P-value < 0.025P-value < 0.005

Sketch the sampling distribution and show the area corresponding to the P-value.

(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 α?

At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.    At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

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

There is sufficient evidence at the 0.05 level to conclude that more than 35% of the students have jobs.There is insufficient evidence at the 0.05 level to conclude that more than 35% of the students have jobs.

Answer 1

Solution :

the level of significance = 0.05

This is the right tailed test .

The null and alternative hypothesis is

standard normal, since np > 5 and nq > 5.

H₀ : p =0.35

H_a : p > 0.35

$\hat p$ = x / n = 40 / 85 = 0.4706

Test statistic = z

= $\hat p$ - P₀ / [$\sqrt$P_{0 *} (1 - P₀ ) / n]

= 0.4706 - 0.35 / [$\sqrt$(0.35 * 0.65) / 85]

= 2.33

0.025 < P-value < 0.005

(d)

At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.

(e)

There is sufficient evidence at the 0.05 level to conclude that more than 35% of the students have jobs.