Question

Homework/ Class Exercise on Topic 6 to 8 Goodyear Tires Malaysia which located at Subang Jaya currently produces tires at their plant during two 12-hour shifts. The night-shift employees are planning to ask for a raise because they believe they are producing more tires per shift than the day-shift. The production supervisor randomly selected some daily production runs from each shift with the results given below (in 1,000s of tires produced). a) Shift Production (in 1,000s) Day | 107.5 | 118.6 | 124.6 | 101.6 | 113.6 | 119.6 ļ 1206 Tì09. | 105,9 Night 115.6 1094 121.6 1287 136.6 125.4 121.3 180.6 17.5 At 5 percent significance level, do these data indicate that the night shift is producing more tires per shift?

Answer 1

H0: Null Hypothesis: $\mu 1\geq \mu 2$

HA: Alternative Hypothesis: $\mu 1<\mu 2$

From the given data, the following statistics are calculated:

n1= 9

$\bar{x}$1 = 113.51

s1 = 7.8058

n2 = 9

$\bar{x}$2 = 128.52

s2 = 21.0397

$s_{P}=\sqrt{\frac{(n1-1)s1^{2}&plus;(n2-1)s2^{2}}{n1&plus;n2-2}}$

$=\sqrt{\frac{(9-1)\times 7.8058^{2}&plus;(9-1)\times 21.0397^{2}}{9&plus;9-2}}=\sqrt{251.8}=15.8682$

$SE=s_{P}\sqrt{1/n1&plus;1/n2}$

$=15.8682\sqrt{1/9&plus;1/9}=7.4803$

Test Statistic is:

t = (113.51 - 128.52)/7.4803 = - 2.0066

$\alpha$ = 0.05

ndf =9 +9 - 2 = 16

One Tail - Left SideTest

From Table, critical value of t = - 1.7459

Since the calculated value of t = - 2.0066 is less than critical value of t = - 1.7459, the difference is significant. Reject null hypothesis.

Conclusion:

The data support the claim that the night shift is producing more tires per shift.