Question

an 5 Sout of 5 points Two suppliers manufacture a pagar used in a laserprer. The mpact strength of these pars measured in to pounds is an important character. A random sample of 10 years from supplier results in a man of 2.0 and a samples of 22.5 and another random sample of 16 gears from the second supplier results in a mean of 21.50 and samples of Do the data support the claim that the mean impact strength of gears from supplier is at least foot poundsigher than that of the tests and that both populations are normaldruted but the variances are equal Use a decat

Answer 1

Here we have two different manufactures for plastic gear. The comparison is between the strength. Since the samples are independent , assumed to be normally distributed and the pop SD are not given, we will use independent samples t-test for difference of population means. We assume that the population variances are equal.

	Gear (1)	Gear (2)
Mean	289.30	321.5
Variance	506.25	441.
SD	22.5	21
n	10	16

The claim is see if gear (2) is atleast (25 or more than 25) higher in strength than gear 1. (2) > (1) + 25 : (1) - (2) < -25. This is one left tailed test.

$H_{0}:\mu1-\mu2=-25$

$H_{1}:\mu1-\mu2<-25$

Pooled variance =

Vari(nı – 1) + Varz(n2 - 1) ni + n2 - 2

= 465.4688

Test Stat =

(21-12) – (ui - U2) Pooled Var(1/n1 + 1/n2)

The null difference = -25

Test stat = -0.8279