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.
