Based on an indication that mean daily car rental rates may be higher for Boston than for Dallas, a survey of eight car rental companies in Boston is taken and the sample mean car rental rate is $47, with a standard deviation of $3. Further, suppose a survey of nine car rental companies in Dallas results in a sample mean of $44 and a standard deviation of $3. Use alpha = 0.05 to test to determine whether the average daily car rental rates in Boston are significantly higher than those in Dallas. Assume car rental rates are normally distributed and the population variances are equal. The null hypothesis for this problem is . a) μ1 - μ2 < 0
b) μ1 - μ2 > 0
c) μ1 - μ2 = 1
d) μ1 - μ2 ≠ 0
e) μ1 - μ2 = 0

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. Let μ1 be the mean daily car rental rates for Boston and μ2 be the mean daily car rental rates for Dallas.

The random variable is μ1 - μ2 = difference in the mean daily car rental rates for Boston and the mean daily car rental rates for Dallas

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 > μ2 H1 : μ1 - μ2 > 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

From the information given,

μ1 = 47

μ2 = 44

s1 = 3

s2 = 3

n1 = 8

n2 = 9

t = (47 - 44)/√(3²/8 + 3²/9)

t = 1.41

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [3²/8 + 3²/9]²/[(1/8 - 1)(3²/8)² + (1/9 - 1)(3²/9)²] = 4.515625/0.28571428571

df = 16

We would determine the probability value from the t test calculator. It becomes

p value = 0.09

Since alpha, 0.05 < than the p value, 0.09, then we would fail to reject the null hypothesis.

16.9 times 2.85

step-by-step explanation: