statistics , two different results related in their standard deviation

Question

I have a question I don't know how to think about it. We have a sample of $25$ people, $16$ of them smoke and $9$ don't. The average capacity of their lungs in smokers is $103$ and in non-smokers is $95$ and the standard deviation of all people in the sample is $10$. Can we be sure with $95\%$ confidence that smokers have statistically more lung capacity? My problem is that I can't understand basically what does that deviation mean and how can it help? The problem seems easy but I am so confused about it. Thank you.

Answer 1

I think this must be a 2-sample z-procedure, because you have two samples (smokers and nonsmokers) and the common population standard deviation is given as $\sigma = 10.$ Please see if you can find the formulas I use below in your text (or lecture notes), and read any derivation or intuitive justification that may be provided there.

One-sided z-test. The null hypothesis that smokers have the same lung capacity as nonsmokers. The one-sided alternative is that smokers have greater lung capacity.

You would reject the null hypothesis (and suppose that smokers have statistically greater greater lung capacity) if the statistic

$$ Z = \frac{\bar X_s - \bar X_n}{\sigma \sqrt{1/n_s + 1/n_n}} \ge 1.645.$$ With $\bar X_s = 103,\;\bar X_n = 95$ and $n_s = 16,\;n_n = 9,\;$ I computed $Z = 1.92.$ So we do reject the null hypothesis. This is a test at the 5% significance level; 1.645 cuts 5% of the area from the upper tail of a standard normal distribution.

One-sided z-interval. Corresponding to this one-sided test of hypothesis is the one sided 95% confidence interval for the $\mu_s - \mu_n$ difference in population means. The lower bound on this difference is $$ \bar X_x - \bar X_n - 1.645 \sigma \sqrt{\frac{1}{n_s}+\frac{1}{n_n}},$$ and this bound computes to 1.146.

In other words, subject to the assumption of random normal data and knowing $\sigma = 10$, we have 95% confidence that the lung capacity of smokers is at least 1.146 units better than the lung capacity of nonsmokers. Compared to mean measurements near 100, a difference of less than 2, even if significant, has to be considered of 'borderline' practical importance, at best.

Brief comments on the role of $\sigma:\;$ In the smoking group, we have $X_1, \dots, X_{16}$ independent and distributed as $Norm(\mu_s, \sigma = 10).$ This implies $\bar X_s \sim Norm(\mu_s, \sigma/\sqrt{n}).$ Similarly for nonsmokers. Then the variance of the difference of the two independent sample means is $$Var(\bar X_s - \bar X_n) = \sigma^2/n_s + \sigma^2/n_n.$$ Notice that variances are added, not subtracted. Then $$SD(\bar X_s - \bar X_n) = \sigma\sqrt{1/n_s + 1/n_n},$$ which plays a role in the formulas above.

Note: A two-sided test would not quite reject, and a two-sided confidence interval would barely include 0. So the particular conclusions above are based on the "one-sided" language in the question: whether smokers have "statistically more lung capacity" (as opposed to "statistically different lung capacity).