Question

a)

The average salary in this city is $43,000. Is the average less for single people? 70 randomly selected single people who were surveyed had an average salary of $41,011 and a standard deviation of $8,900. a) What is the Null and alternative hypotheses. b) What can be concluded at the a = 0.05 level of significance? c) In words what is your final conclusion. If n=132, 7 = 34, and s = 9, find the margin of error at a 99% confidence level.

Answer 1

a)

Null Hypothesis H0: $\mu$ = $43,000

Alternative Hypothesis H1: $\mu$ < $43,000

b)

Sample mean, = $41,011

T

Sample standard deviation s = $8900

Since we do not know the true population standard deviation we will conduct one sample t test.

Standard error of mean, SE = s /
n
= 8900 / $\sqrt{70}$ = 1063.753

Test statistic, t = (
T
- $\mu$ ) / SE =  (41011 - 43000) / 1063.753 = -1.87

Degree of freedom = n-1 = 70-1 = 69

p-value = p(t < -1.87, df = 69) =  0.0329

Since, p-value is less than 0.05 significance level, we reject null hypothesis H0 at 5% level of significance.

(c)

There is sufficient evidence at 5% significance level to conclude that the true mean salary for single people in the the city is less than $43,000

Standard error of mean = s /
n
= 9 / $\sqrt{132}$ = 0.7833495

Degree of freedom = n-1 = 132-1 = 131

Critical value of t at 99% confidence level and df = 131 is 2.61

Margin of error = t * SE = 2.61 * 0.7833495 = 2.0445