0
$\begingroup$

As part of the quality-control program for a catalyst manufacturing line, the raw materials (alumina and a binder) are tested for purity. The process requires that the purity of the alumina be greater than 85%. A random sample from a recent shipment of alumina yielded the following results (in %): $$ 93.2, 87.0, 92.1, 90.1, 87.3, 93.6$$ Using that information, I get that $$n = 6, \bar X =.91, S = 0.03$$ A) State the appropriate null and alternative hypotheses. I have that $$H_0:\mu_0<.85\space H_1:\mu_0\le.85$$ B)Compute the p-value. $$test\space statistic\space t=\frac{\sqrt{n}(\bar X - \mu_0)}{\sigma}=\frac{\sqrt{6}(.91-.85)}{.03} =4.9$$ Since n is sufficiently small, we calculate the t-distribution Assuming a 5% significance level, and with 5 degrees of freedom $t_{.025,5} =2.571$ C)Should the shipment be accepted? Explain. The shipment should not be accepted because 2.571< 4.9 which disproves the null hypothesis. We therefore accept the alternative hypothesis

$\endgroup$
0
$\begingroup$

$a)$: The null and alternative hypothesis should be:

$H_0: \mu = 0.85, H_1: \mu > 0.85$

$b)$: It is not necessary to get an exact $P$-value, you can estimate it with $df = n-1 = 6-1 = 5, \alpha = 0.05$, and the test is a one-tail test ( right tail ), then $P$-value $< 0.005 < 0.05 = \alpha $ because $4.9 > 4.032$ which corresponds to the $\alpha = 0.005$ using Triola textbook. Thus we reject $H_0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.