Question

A survey found that the average daily cost to rent a car in Los Angeles is $102.24 and in Las Vegas is $97.35. The data were collected from two random samples of 40 in each of the two cities and the population standard deviations are $5.98 for Los Angeles and $4.21 for Las Vegas. At the 0.05 level of significance, construct a confidence interval for the difference in the means and then decide if there is a significant difference in the rates between the two cities. Let the sample from Los Angeles be Group 1 and the sample from Las Vegas be Group 2.Confidence Interval (round to 4 decimal places):
< mu1 - mu2 <
Is there a significant difference in the means?

Answer 1

Answer:

Step-by-step explanation:

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

Where

x1 = average daily cost to rent a car in Los Angeles

x2 = average daily cost to rent a car in Las Vegas

s1 = sample standard deviation for Los Angeles

s2 = sample standard deviation for Las Vegas

n1 = number of sampled cars in Los Angeles

n2 = number of sampled cars in Las Vegas

Degree of freedom = (n1 - ) + (n2 - 1) = (40 - 1) + (40 - 1) = 38

For a 95% confidence interval, the t score from the t distribution table is 2.024

From the information given,

x1 = 102.24

s1 = 5.98

n1 = 40

x2 = 97.35

s2 = 4.21

n2 = 40

x1 - x2 = 102.24 - 97.35 = 4.89

Margin of error = z√(s1²/n1 + s2²/n2) = 2.024√(5.98²/40 + 4.21²/40) = 2.024√1.3371125

= 2.34

The 95% confidence interval is 4.89 ± 2.34

Hypothesis testing

This is a test of 2 independent groups. The population standard deviations are not known. Let μ1 be the mean average daily cost to rent a car in Los Angeles and μ2 be the the mean average daily cost to rent a car in Las Vegas

The random variable is μ1 - μ2 = difference in the mean average daily cost to rent a car in Los Angeles and the mean average daily cost to rent a car in Las Vegas

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

This is a two tailed test

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)

t = (102.24 - 97.35)/√(5.98²/40 + 4.21²/40)

t = 4.23

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 = [5.98²/40 + 4.21²/40]²/[(1/40 - 1)(5.98²/40)² + (1/40 - 1)(4.21²/40)²] = 1.78786983766/0.02552804373

df = 70

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

p value = 0.00007

Since alpha, 0.05 > than the p value, 0.00007, then we would reject the null hypothesis. Therefore, at 5% significance level, there is sufficient evidence to conclude that there is a significant difference in the rates between the two cities.