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,$H_1:\mu>1800$
- 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 $$Z=\frac{\bar x-\mu}{\frac{s}{\sqrt n}}=\frac{1950-1800}{\frac{500}{\sqrt 40}}=1.897$$$$Z=\frac{\bar x-\mu}{\frac{s}{\sqrt n}}=\frac{1950-1800}{\frac{500}{\sqrt 40}}=1.897,$$
and the rejection region is $Z>1.96$.
Conclusion: We fail to reject null hypothesis.
But here,they they have considered a $t-$$t$-test. But the sample size is large enough to do a $Z-$$Z$-test.
Again, the link doesn't consider upper tail'upper tail'. So theretheir conclusion is in contradiction ofwith mine.
Which one is correct? Which one is correct?
