help with stats homework.

Question

.    Suppose that the percentage of smokers among American adults was 70% in 1954.  We are interested in knowing if the percentage of smokers has changed since then.  We sample 100 American adults and find that 58 of them smoke. Given the information up to this point, what percentage of American adults did not smoke in 1954? In your sample of 100 people, how many did not smoke? Do the appropriate hypothesis test (include all of the usual information).  Use alpha = .01.

 

Answer 1

In 1954, 30% were non-smokers (100-70), while in the recent sample 42% were non-smokers (100-58).

The null hypothesis H₀ is that there is no change to the mean μ=30% (non-smokers). Using the 1954 figures the variance, σ²=30×0.70=21, and the standard deviation is √21=4.6% approx. So we estimate that the percentage of non-smokers in 1954=30±4.6 (or the percentage of smokers is 70±4.6). The alternative hypothesis H_A is that the new mean for non-smokers is significantly different from the 1954 figure of 30%.

The difference in the means for non-smokers is 42-30=12%. Using the population standard deviation from 1954, we can work out a Z-score=12/4.6=2.61 approx, corresponding to a p-value of 0.9955. The given significance level is 0.01, but for a 2-tail test we split this into 0.005 for each tail which gives us the confidence interval 99%. The p-value is slightly bigger than 0.995 or 99.5%, which means we should reject H₀ in favour of H_A. We conclude that the mean percentage of non-smokers is different from what it was in 1954. If H_A had been that there was a rise in the mean percentage of non-smokers we would have been comparing the p-value with 99% instead of 99.5%, so the results would have been even more significant.