Question

A poll conducted in 1971 asked 1009 people, During the past year about how many books either hardcover or paperback did you read either all or part of the way through results of the survey indicated that x=18.9 books and s=19.9 books.
(a) construct a 95% confidence interval for the mean number of books read either all or part of during the preceding year interpret the interval.
(b) compare these results to a recent survey of 1009 people. The results of the survey indicate that x = 12.5 books and s= 15.7 books. A 95% confidence interval for this survey is( 11.53, 13.47). Were people reading more in 1971 than they are today?

Answer 1

a)

CI for = 95%

n = 1009

mean = 18.9

z-value of 95% CI = 1.9600

std. dev. = 19.9

SE = std.dev./sqrt(n) = 19.9/sqrt(1009) = 0.62648

ME = z*SE = 1.96*0.62648 = 1.2279

Lower Limit = Mean - ME = 18.9 - 1.2279 = 17.67212

Upper Limit = Mean + ME = 18.9 - 1.2279 = 20.12788

95% CI (17.6721 , 20.1279)

One can be 95% confident that true mean of population reading number of books in 1971 lies in the interval

b)

As there is no overlap in the two intervals, we can conclude that the people were reading more books in 1971