Question

Randomly selected 8080 student cars have ages with a mean of 88 years and a standard deviation of 3.6 years, while randomly selected 9595 faculty cars have ages with a mean of 5.4 years and a standard deviation of 3.7 years.
1.    Use a 0.03 significance level to test the claim that student cars are older than faculty cars.
The test statistic is  
The critical value is  
Is there sufficient evidence to support the claim that student cars are older than faculty cars?

A. Yes
B. No

2.    Construct a 97% 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. _______ <(?1−?2)< _____

Answer 1

Solution:-

1)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁< u₂
Alternative hypothesis: u₁ > u₂

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.03. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 0.55327
DF = 173

t = [ (x₁ - x₂) - d ] / SE

t = 4.69

t_critical = + 1.894

Rejection region is t > 1.894

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is thesize of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

Interpret results. Since the t-value (4.69) is greater than the t critical(1.894), hence we have to reject the null hypothesis.

Yes, there is sufficient evidence to support the claim that student cars are older than faculty cars.

2) 97% confidence interval estimate of the difference μ₁ − μ₂ is C.I = ( 1.389, 3.811).

C.I = 2.6 + 2.188 × 0.55327

C.I = 2.6 + 1.2106

C.I = ( 1.389, 3.811)