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An insurance company is reviewing its current policy rates. When originally setting the rates they believed that the average claim amount was $1,800$. They are concerned that the true mean is actually higher than this, because they could potentially lose a lot of money. They randomly select 40 claims, and calculate a sample mean of $1,950$. Assuming that the standard deviation of claims is $500$, and set $\alpha = 0.05$, test to see if the insurance company should be concerned.

My attempt:

  • Null hypothesis: $H_0:\mu\leq1800$
  • Alternative hypothesis: $H_1:\mu>1800$

Since our sample size is large, we will do $Z$-test. The test statistic is $$Z=\frac{\bar x-\mu}{\frac{s}{\sqrt n}}=\frac{1950-1800}{\frac{500}{\sqrt 40}}=1.897,$$

and the rejection region is $Z>1.96$.

Conclusion: We fail to reject null hypothesis.

But here, they have considered a $t$-test. But the sample size is large enough to do a $Z$-test.

Again, the link doesn't consider 'upper tail'. So their conclusion is in contradiction with mine.

Which one is correct?

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    $\begingroup$ The test is one-sided, you need to check the rejection region of your approach. $\endgroup$ – QuantIbex Aug 10 '13 at 12:55
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    $\begingroup$ If the population standard deviation is known, then you can use a $z$-test. As far as I understand your setup, you assume to know the population standard deviation to be $500$. If you estimate the population standard deviation by the sample standard deviation, a $t$-test is normally used. In large samples, the difference between the $z$-test and $t$-test are negligible. The critical $t$ value on the linked page is the $0.95$ quantile of the $t$-distribution with $34$ degrees of freedom. The $0.95$ quantile of the standard normal is $1.644$, not $1.96$ (because you have a one-sided test). $\endgroup$ – COOLSerdash Aug 10 '13 at 12:57
  • $\begingroup$ @COOLSerdash Can you please give me reference on one-sided test and two-sided test. My main problem is to compute the value based on one-sided test and two-sided test. $\endgroup$ – ABC Aug 10 '13 at 13:28
  • $\begingroup$ For a two sided test I believe you just use alpha/2 as alpha. $\endgroup$ – Justin Bozonier Aug 10 '13 at 13:41
  • $\begingroup$ @JustinBozonier A two sided test, statistic $z=2.42$. $p-$value$=2*0,0078$. Can you please explain why did we multiply $2$ instead of dividing? $\endgroup$ – ABC Aug 10 '13 at 13:48
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"They randomly select 40 claims..."

First, you only have 40 samples. (not 40 samples!!, One sample with 40 cases). That's not enough to do a Z-test. You are in solid T-test territory as is your example.

"They are concerned that the true mean is actually higher than this..." So this will be a one sided test. Looking up the test statistic for a one sided t-test with 39 degrees of freedom (from here http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf)

Also according to that table you don't have a z statistic until you've had over 1,000 samples.

The closest I can come is 40 so the statistic would be 1.684. Your null and alternative hypotheses look good to me

Using the one sample T-test referenced here (the same as your z-statistic): http://en.wikipedia.org/wiki/Student's_t-test

$$t = \frac{\overline x - \mu_0 }{\frac{s}{\sqrt n}} = \frac{1,950 - 1,800 }{\frac{500}{\sqrt 40}} = 1.8974$$

Which is the same value as you. Since $t > \alpha$ we reject the null hypothesis.

I think the main issue here is using a one sided test vs two sided and understanding you are not able to use a z-test with so few samples.

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