Answer to Question #214625 in Statistics and Probability for Lyndk85

Question

A hospital is studying the delivery time of two different laundry companies. The hospital has been using company a for the past year and is basically satisfied with the time the company takes to return laundry to the hospital. The hospital is prepared to stay with company a if the delivery time is the same or less than that of competitor company, Company B. However, if the hospital finds that the mean delivery time of company b is less than that of company A, it will start using the laundry services of company B. Independent samples show the following delivery time characteristics for both companies:

company A

n1 = 20

mean x1 = 10 days

s1 = 3 days

company b

n2 = 30

mean x2 = 12.5 days

s2 = 2 days

what are the null and alternative hypothesis for this situation?

with an a=.25 what is your conclusion for the hypothesis from part a?

what action do you recommend in terms of which laundry company the hospital should contract?

Answer 1

$H_0:\mu_1=\mu_2$

$H_a:\mu_1<\mu_2$

First we test the equality of variances.

$H_0:\sigma_1^2=\sigma_2^2$

$H_a:\sigma_1^2\ne\sigma_2^2$

$F=\frac{s_1^2}{s_2^2}$

$=\frac{9}{4}$

$=2.25$

$cv=F_{0.05,19,29}=1.958$

$2.25>1.958$, thus, we reject the null hypothesis. Variances are significantly different.

(NOTE)

Assuming unequal variances can be used under two conditions;

1. When you know that the variances are unequal or
2. When you do not know whether variances are equal or not

Thus, it is not always required to conduct equality of variance test. If the variances are equal the equal and unequal variances versions of the t-test yield similar results (even when the sample sizes are unequal).

The samples are independent with unequal variances(sample variance for company A is 9, while sample variance for company B is 4); thus, two independent samples t-test assuming unequal variances is appropriate for this test.

$t = \frac{\bar{X_{1}} - \bar{X_{2}}} {\sqrt{\frac{{s^{2}_{1}}}{n_{1}} + \frac{{s^{2}_{2}}}{n_{2}}}}$

$df = \frac{(\frac{s^{2}_{1}}{n_{1}} + \frac{s^{2}_{2}}{n_{2}})^{2}} {\frac{(\frac{s^{2}_{1}}{n_{1}})^{2}}{n_{1}-1 }+ \frac{(\frac{s^{2}_{2}}{n_{2}})^{2}}{n_{2}-1 }}$

$t = \frac{10 - 12.5} {\sqrt{\frac{{9}}{20} + \frac{{4}}{30}}}$

$=-3.273$

$df = \frac{(\frac{9}{20} + \frac{4}{30})^{2}} {\frac{(\frac{9}{20})^{2}}{19 }+ \frac{(\frac{4}{30})^{2}}{29 }}$

$=30.19$ (approximately =30)

Critical value $t_{0.25,30}=-0.6828$ (Left tailed)

The absolute value of the test statistic (3.273) is greater than the absolute critical value (0.6828), thus, we reject the null hypothesis. There is sufficient evidence to support the claim that the average delivery time for company A is lower than the average delivery time for company B. The hospital should contract company A.