Question

In order to estimate the difference between the MEAN yearly incomes of marketing managers in the East and West of the United States, the following information was gathered.

East	West
n = 40	N = 42
X = $73,000	X= $71,500
S = 3500	S= 4000

1. Develop an interval estimate for the difference between the MEAN yearly incomes of the marketing managers in the East and West. Use α = 0.05. Note that the population standard deviation is unknown, but the sample standard deviation is known. You will also need to calculate the degrees of freedom as shown on page 434 so that you can calculate t...
2. You will reuse a lot of what you did in the problem above. Use the p-value approach to conduct a full hypothesis test to determine if the MEAN yearly income of marketing managers in the East is significantly different from the West. p. 435 can help. Your difference in your formula is 0.

Answer 1

Answer)

1)

As the population s.d is unknown we will use t table to construct the interval

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

= 39

For df 39 and test 95% confidence level (as confidence level is = 1 - alpha = 1-0.05 = 0.95 = 95%) critical value t is = 2.023

Margin of error (MOE) = t*standard error

t = 2.023

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

S1 = 3500, S2 = 4000

N1 = 40, N2 = 42

After substitution

MOE = 1677.01987850

Confidence interval is given by

(X1-x2)-MOE < (u1-u2) < (X1-X2)+MOE

x1 = 73000, x2 = 71500

−177.01987850 < (u1-u2) < 3177.01987850

2)

Ho : u1 = u2

Ha : u1 not equal to u2

Test statistics = (x1-x2)/standard error

Test statistics = 1.809

For df 39 and 1.809 test statistics

P-value from t distribution is = 0.078165

As the obtained p-value is greater than the given significance level, (0.05)

We fail to reject the null hypothesis

And we do not have enough evidence to support the claim that there is a significant difference between the means