Randomly selected 150 student cars have ages with a mean of 7.3 years and a standard deviation of 3.6 years, while randomly selected 65 faculty cars have ages with a mean of 5.9 years and a standard deviation of 3.5 years. 1. Use a 0.01 significance level to test the claim that student cars are older than faculty cars.
The test statistic is
Is there sufficient evidence to support the claim that student cars are older than faculty cars?

A. Yes
B. No

2. Construct a 99% confidence interval estimate of the difference μ1−μ2, where μ1 is the mean age of student cars and μ2 is the mean age of faculty cars. <(μ1−μ2)

1) B. NO

2) -0.049 < (μ1 − μ2) > 2.751

Step-by-step explanation:

Data of student cars;

Sample size; n1 = 150

Sample mean; x1¯ = 7.3

Sample standard deviation; s1 = 3.6

Faculty cars data;

Sample size; n2 = 65

Sample mean; x2¯ = 5.9

Sample standard deviation; s2 = 3.5

The hypotheses is defined as;

Null hypothesis; H0: μ1 = μ2

Alternative hypothesis; Ha: μ1 > μ2

1) Formula for test statistic for 2 sample test is;

z = (x1¯ - x2¯)/√[((S1)²/n1) + ((S2)²/n2)

z = (7.3 - 5.9)/√[((3.6)²/150) + ((3.5)²/65)

z = 1.4/√((12.96/150) + (12.25/65))

z = 1.4/√0.2749

z = 1.4/0.5243

z = 2.67

From online p-value from z-score calculator attached, using z = 2.67, one tailed test, significance level = 0.01,we have;

P-value = 0.003793

This is less than the significance level, thus we will reject the null hypothesis and conclude that there is no sufficient evidence to support the claim that student cars are older than faculty cars.

2) Formula for the confidence interval is;

CI = μ1−μ2 = (x1¯ - x¯2) ± z√[((S1)²/n1) + ((S2)²/n2)

At significance level of 99%, z-score value is 2.576

Thus;

μ1−μ2 = (7.3 - 5.9) ± 2.576√[((3.6)²/150) + ((3.5)²/65)

μ1 − μ2 = 1.4 ± 2.576(0.5243)

-0.049 < (μ1 − μ2) > 2.751