Statistic worksheet

Question

1.     Of a random sample of 198 marketing students 98 rated a case of resume inflation as unethical.

Based on this information, a statistician computed for the population proportion a confidence interval extending from 0.445 to 0.545. What is the confidence level of this interval?

2.  A college placement officer wants to determine whether male and female business graduates receive, on average, different salary offers for their first position after graduation. The placement officer randomly selected eight pairs of business graduates in such a way that the qualifications, interests, and backgrounds of the members of any pair were as similar as possible. The major difference was that one member of each pair was male and one was female. The accompanying table shows the highest salary offer received by each sample member at the end of the recruiting round. Assuming that the distributions are normal, test the null hypothesis that the population means are equal against the alternative that the true mean for males is higher than for females. Use α=0.10. Interpret your result.

Pair	Highest Salary Offer (In Dollars)
	Male	Female
1	26,200	22,600
2	24,700	23,600
3	28,400	29,300
4	21,700	22,300
5	28,600	26,200
6	29,300	25,900
7	28,300	28,500
8	24,300	21,300

3.  An instructor has decided to introduce a greater component of independent study into an intermediate macroeconomics course, as a way of motivating students to work independently and think more carefully about the course material. A colleague cautions that a possible consequence may be increased variability in student performance. However, the instructor responds that she would expect less variability. From her records, she found that in the past, student scores on the final exam for this course followed a normal distribution with standard deviation 18.2 points. For a class of twenty five students using the new approach, the standard deviation of scores on the final exam was 15.3 points. Assuming that these twenty five students can be viewed as a random sample of all those who might be subjected to the new approach, test the null hypothesis that the population standard deviation is at least 18.2 points against the alternative that it is lower. Use α=0.05. Find a p-value. Interpret your results.

Thanks for helping me i cant solve those problems...

Answer 1

1.) p = 98/198 = 0.495 p-hat = 0.545

We will use the z-test for 2 different proportions.

z = (0.545 - 0.495) / √(0.495)(0.505)/198 = 0.05/0.03553 = 1.41

P(z < 1.41) = 0.92. Our confidence level for this interval is 92%.

2.) We will use the t-test for 2 different means.

H_o: μ_male = μ_female

H_A: μ_male > μ_female

Male: n = 8, x-bar = $26,437.50, s = $2680.05

Female: n = 8, x-bar = $24,962.50, s = $2968.37

I use the excel to calculate the sample mean and sample standard deviation in the data analysis.

t = (26437.5 - 24,962.5) / √(2680.05)²/8 + (2968.37)²/8 = 1475/1414.05 = 1.04

d.f. = 8 - 1 = 7 α = 0.10 t_0.10,7 = 1.415 (use the t-distribution table to find the critical value)

Since 1.04 < 1.415, we failed to reject the null hypothesis. There is no sufficient evidence to show that the average salary for men is higher than the average salary for women.

3.) We will the chi-squared test to compare standard deviation.

H_o: σ ≥ 18.2

H_A: σ < 18.2

χ² = (25-1)(15.3)² / (18.2)² = 16.961

χ²_0.005 = 36.42 (with degrees of freedom = 24) Use the chi-squared table to find the critical value.

Since 16.961 < 36.42 and the p-value is 0.850321 > 0.05, we fail to reject the null hypothesis.

Answer 2

Remember a confidence interval is an estimate of an unknown population parameter, a "best guess" plus or minus a margin of error. Since we are using the results of a survey to estimate the proportion of all students who believe that resume inflation is unethical, our "best guess" is the sample proportion (p-hat), 98/198=.495. We are given the interval, so we can calulate the margin of error. Our margin of error it is the distance from the center of the interval to the edge of the interval; (.545-.445) /2=0.05. Check it. Add and subtract 0.05 from .495.

The formula we use to calculate the margin of error for a confidence interval for a population proportion is

error = z* sqrt(p-hat(1-p-hat)/n)

We simply need to rewrite this equation to solve for z*, the critical value.

z*= error/ [sqrt(p-hat(1-p-hat)/n)]=.05/[sqrt(.495(1-.495)/198]=1.41

We have our critical value, but remember a confidence interval is two-sided. To determine our level of confidence, we need to consider both the area to the right of z=1.41 and the area to the left of z=-1.41. From the normal tbles we can see that this area is about 16%. So our confidence level is 84%.