2
$\begingroup$

Experience in investigating insurance claims shows that the average cost to process a claim is approximately normally distributed with a mean of 80 dollars. New cost-cutting measures were started and a sample of 25 claims was tested. The sample mean of the costs to process these claims was 76% and the sample standard deviation of the costs was $10. We would like to test whether the cost-cutting measures seem to be working at the 5% significance level. Find the critical value for this test.

I need help please.

I have calculated that

$H_0: \mu = 80$ is the null hypothesis

$H_1: \mu < 80$ is the alternative hypothesis

Also the test statistics i have calculated is like

$z = (76 - 80)/(10/\sqrt{25})=-2$

Can some one help me now in finding the critical value so I can conclude which hypothesis is true here. Thanks

$\endgroup$
0
$\begingroup$

You are basically there. I assume that you are doing a $Z$ test here (although if you have been taught how to perform a $t$-test, then that test, with 24 degrees of freedom, might be more appropriate).

You want to know if your $Z$ statistic of $-2$ is significant. Since you are performing a one tailed test, the critical value for this statistic is the $\alpha$-quantile of the $Z$ distribution (i.e., the standard normal distribution). The critical value you use would depend on the significance level of the test. If you want to perform the test at the $99\%$ significance level, then the critical value is $-2.326$. The rejection region is therefore the set $(-\infty,-2.326)$. If you want to conduct the test at the $95\%$ significance level, the critical value is $-1.645$. So, since the observed value of $Z$ is $-2$, you would retain the null hypothesis if you wanted to work at the $99\%$ significance level, and you'd reject it if you wanted to work at the $95\%$ level.

I got these values from an online normal distribution calculator, but you can get them from any table of normal values too (although maybe not to the same level of precision).

Since your decision to reject or retain the null hypothesis depends on the significance level you choose, it isn't really reasonable to say that you have determined which hypothesis is true. You have merely decided whether the observed data are consistent with the hypothesis of no improvement.

(It should also be noted that you are assuming that the distribution of costs is still normally distributed after the cost cutting measures. You did not state explicitly that this was the case, but it is necessary in order to use either the $t$ or $Z$ test.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Sir what degrees of freedom you have used here? $\endgroup$ – Salman Apr 25 '14 at 21:07
  • $\begingroup$ Since you wrote $ Z=\ldots $ and not $ T=\ldots$ when calculating your test statistic, I assumed that you wanted to do a $ Z $ test and not a $ t $ test. This isn't really correct, but people often do $ Z $ tests when first learning this stuff and I didn't want to confuse you, so I just used the normal distribution (which is equivalent to the $ t_{\infty} $ distribution). If you wanted to be more correct, your critical value would be the $\alpha $ quantile of the $ t_{24} $ distribution. $\endgroup$ – Unwisdom Apr 25 '14 at 21:57

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.