Question

(3 points) Randomly selected 21 student cars have ages with a mean of 7.3 years and a standard deviation of 3.4 years, while randomly selected 14 faculty cars have ages with a mean of 5.5 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 (a) The test statistic is 2.32 (b) The critical value is 2.32 (c) Is there sufficient evidence to support the claim that student cars are older than faculty cars? OA. Yes OB. No 2. Construct a 99% confidence interval estimate of the difference μ5-11f, where age of student cars and uf is the mean age of faculty cars is the mean ) s

Answer 1

Answer:

$\ H_0: \mu_1 = \mu_2 \ \ \ H_1: \mu_1 > \mu_2$

Upper tail test

1).

a).Test statistic = 0.6741

b). Critical value =2.4448

c).

No.

2).

99% CI for difference = -1.4439 < difference < 5.0439.

$t = \frac{\bar{x_1} - \bar{x_2}}{\sqrt{ \frac{(n_1 - 1) s_1^2 &plus; (n_2 - 1) s_2^2} {(n_1 &plus;n_2 -2)} *{(\frac{1}{n_1} &plus; \frac{1}{n_2})}}}$

Pooled-Variance t Test for the Difference Between Two Means
(assumes equal population variances)
Data
Hypothesized Difference	1
Level of Significance	0.01
Population 1 Sample
Sample Size	21
Sample Mean	7.3
Sample Standard Deviation	3.4
Population 2 Sample
Sample Size	14
Sample Mean	5.5
Sample Standard Deviation	3.5
Intermediate Calculations
Population 1 Sample Degrees of Freedom	20
Population 2 Sample Degrees of Freedom	13
Total Degrees of Freedom	33
Pooled Variance	11.8318
Standard Error	1.1868
Difference in Sample Means	1.8000
t Test Statistic	0.6741
Upper-Tail Test
Upper Critical Value	2.4448
p-Value	0.2525
Do not reject the null hypothesis

$CI = ( \bar x_1 - \bar x_2) \pm t*se$

Confidence Interval Estimate
for the Difference Between Two Means
Data
Confidence Level	99%
Intermediate Calculations
Degrees of Freedom	33
t Value	2.7333
Interval Half Width	3.2439
Confidence Interval
Interval Lower Limit	-1.4439
Interval Upper Limit	5.0439