Do doctors have a shorter lifespan than general population with this data:

Known lifespan for US subject in last $2$ decades = $70.1$, with a standard deviation of $7.6$.

A sample from $51$ doctors from the above population has a mean lifespan of $65.7$ with a standard deviation of $6.8$. Use $\alpha = 0.05$.

(1) compare $\alpha$ to $p$-value.

(2) What is Null Hypothesis and should it be rejected? Use the $t$ test for this problem.

So, right off the bat, I don't know why this problem would say use the $t$ test when $n$ is larger than $30$ and I also know the population's standard deviation. I would think I would use the $z$ test. So, confused here.

Secondly, my text seems to indicated that the Alternate Hypothesis in this case would be $\mu$ is less than the general pop and the Null is that it is equal to the general pop. Since this is a one-tailed test, I would think the null would be NOT = general pop but greater than equal to. Maybe this is just semantics.

Any thoughts on how to solve this?


Whether to use a $t$-test or $z$-test depends on whether the standard deviation of the population from which the sample is drawn is known. In your case, you are testing whether the average lifespan $\mu_d$ of doctors is less than the average lifespan $\mu_g$ of the general population; i.e., your hypothesis is $$H_0 : \mu_d = \mu_g, \quad \mathrm{vs.} \quad H_a : \mu_d < \mu_g.$$ Under the assumption of the null hypothesis, you still cannot assume that $\sigma_d = \sigma_g$. Another way to think about it is that if you did use a $z$-test, what is the appropriate $\sigma$ to use? Certainly it is not $\sigma_g = 7.6$ because that is the standard deviation of the lifespan of the general population, whereas the standard deviation for doctors is unknown to us. Therefore, a $t$-test is appropriate, even for relatively large sample sizes: you are estimating $s_d$ from the data in your sample.

For your second question, the alternative hypothesis is the statement that, if true, would require the evidence collected in the sample to indicate with a high degree of confidence that it is true. Your decision with respect to $H_0$ and $H_a$ is always between rejecting $H_0$ or failing to reject $H_0$; that is to say, either the evidence suggests $H_a$ is true with at most a probability of $\alpha$ of being mistaken; or there is insufficient evidence to suggest $H_a$ is true, and the test is inconclusive. At no point do you ever accept $H_0$: the decision always one about $H_a$, which is why $H_a$ is the hypothesis of interest here.

Thus, under the assumption of the null hypothesis, that there is no difference between the average lifespans of the two groups, then you can show that the test statistic $$\frac{\bar x - \mu_g}{s_d/\sqrt{n}} \sim t_{n-1},$$ where $\bar x$ is the sample mean of the doctors' lifespans; $s_d$ is the sample standard deviation; $\mu_g$ is the mean of average lifespans among the general population; and $t_{n-1}$ is the student's $t$-distribution with $n-1$ degrees of freedom. Notice that the population standard deviation $\sigma_g$ is not used.

  • $\begingroup$ Now I see this. Thank you very much! $\endgroup$ – user163862 Jul 28 '14 at 0:16

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