Question

A study is run to estimate the mean total cholesterol level in children 2-6 years of age. A sample of 13 participants is selected and their total cholesterol levels are measured as follows. 170 185 225 240 196 175 180 194 147 223 240 220 210 a) Assume the mean total cholesterol level is 190 in children 2-6 years of age at national level. Test whether the cholesterol level in this sample is significantly different from national level. b) Generate a 95% confidence interval for the true mean total cholesterol levels in children 2- 6 years of age.

Answer 1

Given Sample Valley are 170 185, 225, 240,196,175,180,194.147, 223, 240, 220 2in n = 13 From the above Sample mean Mean () = EX; iš (2005) ū 200.39 Standard deviation (8) is 8 - 2C;})" in-1 Vše C170 -200-39) + (186 -200.39* * - - -4610-200.39) - (4863.08) 2,921.92 8=28.67) 28.67 Sample size Degrees of m n = 13 freedom = 13-1-12. Scanned with CamScanner

Null hypothesis (He) : There is no Signibicant diffedence between cholesterol level of sample and maligral ie, M=190 leve Alternative hypothesie (H: ME190 (claim) This Gorresponds to a two-tailed to test Rejection Region : choose lors fix level of Sighili cance 2=0.08 and the Gritical Valle Gora two failed test is te=2.179 at 12 d.f The Rejection region for this two failed test is R={2:14 > 2.179} Test Statistic : To test the above hypothe sys the test statistic is given by to you a tiny . 200.39-190 * 28.671113 = 1.307 X- ME Under Ho = = 1:307 cs Scanned with CamScanner Top

Decision about null hypothesis Since the test Statistic 1 t1 = 1.307 is less than the tcritical value te 22.179 141 = 1.307 < tc: 2.179 it is then concluded that the null hybothesis is not rejeded. Conclusion: It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the mean to cholesterol of Sample is difter than national level z cs Scanned with CamScanner

interval for the b) : The true 95% Confidence mean 18 G.I = (X-te slon , X+te plon) =(200.39 – 2,179 2 3 , .200:39 + 2.1719 y = (200:30 – 17.31, 200-39 +17.31) (183.09 , 217.70) Therefore the 95% confidence interval if (183.09, 217.70) Scanned with CamScanner