Question

1. The admissions officer at a small college compares the scores on the Scholastic Aptitude Test (SAT) for the school's male and female applicants. A random sample of 15 male applicants results in a SAT scoring mean of 1100 with a standard deviation of 53. A random sample of 5 female applicants results in a SAT scoring mean of 1218 with a standard deviation of 30. Using this data, find the 90% confidence interval for the true mean difference between the scoring mean for male applicants and female applicants. Assume that the population variances are not equal and that the two populations are normally distributed.

- Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

-Find the standard error of the sampling distribution to be used in constructing the confidence interval. Round your answer to the nearest whole number.

-Construct the 90% confidence interval. Round your answers to the nearest whole number.

2.

A steel company is considering the relocation of one of its manufacturing plants. The company’s executives have selected four areas that they believe are suitable locations. However, they want to determine if the average wages are significantly different in any of the locations, since this could have a major impact on the cost of production. A survey of hourly wages of similar workers in each of the four areas is performed with the following results. Do the data indicate a significant difference among the average hourly wages in the three areas?

Hourly Wages ($)

Area 1	Area 2	Area 3
21	10	15
21	24	11
24	18	18
17	22	14
15	22	15
18	16	14
11	24	12
22	21	16

- Find the value of the test statistic to test for a difference in the areas. Round your answer to two decimal places, if necessary.

- Make the decision to reject or fail to reject the null hypothesis of equal average hourly wages in the three areas and state the conclusion in terms of the original problem. Use α=0.0.

Answer 1

1)

As the population standard deviation is not given we will use t distribution to estimate the interval

N1 = 15, x1 = 1100, s1 = 53

N2 = 5, x2 = 1218, s2 = 30

Degrees of freedom is = smaller of n1-1, n2-1 = 4

For 4 dof and 90% confidence level, critical value t from t table is = 2.132

Margin of error (MOE) = t*standard error

Standard error = √{(s1^2/n1)+(s2^2/n2)} = 19.1642027401

Moe = 2.132*19.1642027401 = 40.8580802420

Interval is given by

[(X1-x2)-MOE, (X1-X2)+MOE]

[−158.85808024, −77.141919757]

[-159, -77]