1. A company wants to estimate the average monthly

income of its customers. It randomly selects 100 customers

and asks them to report their income. The sample mean is

$3,500 and the sample standard deviation is $500. What is

the 95% confidence interval for the population mean

income?

a) ($3,401, $3,599)

b) ($3,392, $3,608)

c) ($3,383, $3,617)

d) ($3,374, $3,626)

Answer: c) ($3,383, $3,617)

Rationale: The 95% confidence interval for the population

mean is given by the formula: sample mean ± 1.96 *

(sample standard deviation / square root of sample size).

Plugging in the given values, we get: 3500 ± 1.96 * (500 /

sqrt(100)) = 3500 ± 98. Thus, the confidence interval is

($3,402, $3,598), which is closest to option c.

2. A researcher wants to test whether there is a significant

difference in the average scores of male and female

students on a math test. The researcher randomly assigns

50 male and 50 female students to take the test. The mean

score for males is 75 and the mean score for females is 80.

The pooled standard deviation is 10. What is the value of

the test statistic for a two-sample t-test?

a) -2.5

b) -2.78