Death Statistics Question

Question

A random sample of 100 recorded deaths in the United States during the past year showed an average life span of 71.8 years. Assuming a population standard deviation of 8.9 years, does this seem to indicate that the mean life span today is greater than 70 years? Use a 0.05 level of significance.

Answer 1

Hello Peter,

To answer this question, we test the null hypothesis that the population mean age at death is 70 or younger versus the alternative hypothesis that the population mean is greater than 70 – that is, since the stem of the question asks "does this indicate that mean life span is greater than 70" we want a one-sided alternative hypothesis reflecting greater than 70.

The next question we need to answer is what probability distribution do we base the hypothesis test on? We are not told that the population probability distribution of ages at death is a normal distribution. However, a very important result known as the Central Limit Theorem states that the sampling distribution of the mean will approach a normal distribution as the sample size increases regardless of the population distribution of the random variable, or equivalently in this problem, regardless of whether ages at death follows a normal distribution. I'll explain that a little more simply with an example. Suppose that many students (say 50 students) go out and randomly sample the death file and each obtains a sample of 100 deaths and each calculates the mean age at death for his/her sample. The Central Limit theorem says that these 50 different sample means will follow a probability distribution that is approximately a normal distribution.

So we will use the Central Limit Theorem here to argue that the sample mean follows an approximately normal distribution. The only remaining possible fly in the ointment is that if we did not know the population standard deviation, then we would have to estimate the population standard deviation with the sample standard deviation. Using the sample standard deviation as an estimate of the population standard deviation would imply that we should use the Student's t distribution in the hypothesis test below rather than the normal distribution. However, in this problem, we are given that the population standard deviation is 8.9 years, so we do not need to estimate it, so we will use Central Limit Theorem to imply that our test statistic follows a normal distribution.

Here are the hypothesis test details:

Null hypothesis, H0: population mean age at death is less than or equal to 70 years
Alternative Hypothesis, HA: population mean age at death is greater than 70 years

Test statistic:
z = (sample mean – hypothesized mean) / [standard deviation/(square root of sample size))]
= (71.8-70) / (8.9/square root of 100) = 1.8/ 0.89 = 2.02

Also, as we argued above, the sample mean follows a normal distribution, so that the standardized sample mean (that is, the sample mean minus the population mean divided by the standard deviation of the sample mean) follows a standard normal distribution. In other words, this test statistic follows a standard normal distribution.

Critical value of z at the 0.05 level of significance: 1.645

As explanation, the critical value is the value of the test statistic z such that if the null hypothesis is true, then the probability of obtaining a value of z this large or larger is 0.05. In other words, if the null hypothesis is true, then obtaining a value of z that is greater than or equal to 1.645 would only happen 1 time in 20 samples and we use the rarity of that event as evidence that the null hypothesis is false. Note that for a two tailed test, we would reject the null for either a large positive value of z or a large negative value of z, but here the test is a one tailed test, so we only reject for a large positive value of z.

Decision: Since the value of the test statistic, 2.02, is greater than the critical value 1.645, reject the null hypothesis and conclude that mean age at death is greater than 70 years.

A couple of other quick things to note. To find the critical value, you may be able to use a table of the normal distribution in your statistics text, but since these tables are set up differently in different texts, I will indicate how to do this in Excel. In Excel, use the function NORM.INV(0.95,0,1), that is, just paste "=NORM.INV(0.95,0,1)" into the formula bar in Excel. This function will return the value of z such that there is a 95% chance that z will be less than or equal to this value, which means that obtaining a value larger than this value has only a 5% chance, which is what we want. A second thing to quickly note is that we can find the significance that is associated with a z statistic of 2.02 from the Excel function =NORM.DIST(2.02,0,1,TRUE) – this function tells us the probability of drawing a sample that has a test statistic less than or equal to 2.02 when the null hypothesis is true, and it turns out to have a value of 0.978. Therefore 1 – 0.978 = 0.022 is the probability of obtaining a value as large as or larger than the value 2.02. This means that we would reject the null hypothesis at any significance level numerically greater than 0.022 (e.g., reject at 0.03, 0.04, 0.05 etc.) and not reject the null hypothesis at any significance level numerically less than 0.022 (e.g., not reject at significance levels like 0.02, 0.01, and so on).

Hope this helps. Feel free to ask question of not.

Kind regards,
Hugh