Question

a.) According to a recent​ report, 46% of college student internships are unpaid. A recent survey of 120 college interns at a local university found that 57 had unpaid internships. 1.)Use the​ five-step p-value approach to hypothesis testing and a 0.05 level of significance to determine whether the proportion of college interns that had unpaid internships is different from 0.46. 2.) Assume that the study found that 70 of the 120 college interns had unpaid internships and repeat​ (1). Are the conclusions the​ same? Let π be the population proportion. Determine the null​ hypothesis, H0​, and the alternative​ hypothesis, H1. ​(Type integers or decimals. Do not​ round.)

b.) ​Recently, a large academic medical center determined that 9 of 23 employees in a particular position were male​, whereas 55​% of the employees for this position in the general workforce were male. At the 0.05 level of​ significance, is there evidence that the proportion of males in this position at this medical center is different from what would be expected in the general​ workforce? What are the correct hypotheses to test to determine if the proportion is​ different?

c.) A consulting group recently conducted a global survey of product teams with the goal of better understanding the dynamics of product team performance and uncovering the practices that make these teams successful. One of the survey findings was that 36​% of organizations have a coherent business strategy that they stick to and effectively communicate. Suppose another study is conducted to check the validity of this​result, with the goal of proving that the percentage is less than 36​%. State the null and research hypotheses. Identify the null and alternative hypotheses.

Answer 1

a)1) H₀: P = 0.46

H₁: P 0.46

= 57/120 = 0.475

The test statistic z = ( - P)/sqrt(P(1 - P)/n)

= (0.475 - 0.46)/sqrt(0.46 * (1 - 0.46)/120)

= 0.33

P-value = 2 * P(Z > 0.33)

= 2 * (1 - P(Z < 0.33))

= 2 * (1 - 0.6293)

= 2 * 0.3707 = 0.7414

Since the P-value is greater than the significance level(0.7414 > 0.05), so we should not reject the null hypothesis.

So at 0.05 significance level there is not sufficient evidence to conclude that the proportion of college interns that had unpaid internship is different from 0.46.

2) H₀:   = 0.46

H₁:   0.46

 = 70/120 = 0.583

The test statistic z = ( - )/sqrt((1 - )/n)

= (0.583 - 0.46)/sqrt(0.46 * (1 - 0.46)/120)

= 2.7

P-value = 2 * P(Z > 2.7)

= 2 * (1 - P(Z < 2.7))

= 2 * (1 - 0.9965)

= 2 * 0.0035 = 0.007

Since the P-value is less than the significance level (0.007 < 0.05) , so we should reject the null hypothesis.

So at 0.05 significance level there is sufficient evidence to conclude that the proportion of college interns that had unpaid internship is different from 0.46.

So the conclusions in parts 1 and 2 are different.

B) H₀: P = 0.55

H₁: P 0.55

= 9/23 = 0.3913

The test statistic z = ( - P)/sqrt(P(1 - P)/n)

= (0.3913 - 0.55)/sqrt(0.55 * (1 - 0.55)/23)

= -1.53

P-value = 2 * P(Z < -1. 53)

= 2 * 0.0630 = 0.1260

Since the P-value is greater than the significance level (0.1260 > 0.05), so we should not reject the null hypothesis.

So at 0.05 significance level there is not sufficient evidence to conclude that the proportion of males in this position at this medical center is different from what would be expected in the general workforce.

C) H₀: P = 0.36

H₁: P < 0.36