Question

= 40 adults were selected from A small amount of the trace element selenium, 50-200 micrograms (ug) per day, is considered essential to good health. Suppose that random samples of n, = n, two regions of the United States and that a day's intake of selenium was recorded for each person. The mean and standard deviation of the selenium daily intakes for the 40 adults from region 1 = 140.5 and s, = 17.1 µg. Find a 95% confidence interval for the difference = 167.1 and 24.4 pg, respectively. The corresponding statistics for the 40 adults from region 2 were x, were x. %3D (u, - H,) in the mean selenium intakes for the two regions. (Round your answers to three decimal places.) µg to Hg Interpret this interval. There is a 5% chance that the difference between individual sample means will fall within the interval. In repeated sampling, 5% of all intervals constructed in this manner will enclose the difference in population means. 95% of all differences will fall within the interval. In repeated sampling, 95% of all intervals constructed in this manner will enclose the difference in population means. There is a 95% chance that the difference between individual sample means will fall within the interval. You may need to use the appropriate appendix table or technology to answer this question.

Answer 1

gisen. D = ५० n2=40, K= 167.1, S,=24.4 žg= 140.5, s2=17.1 95% contidence intenval is, C.F:=(X,-X,) + t* (17.1)? 40 (24-4) =(167.1-140.5) t 1.99) 26.6 + 22.1943 %3D = (4.406, 48.794) contridence interval és 4.406 to 48:794 95% of all differences will fall withio Intervo the înterval. Condlusion: