Question

please show work. clearly

round to 3 decimal places

Consider: Sugar content in a particular brand of cranberry sauce. A random sample was selected in order to estimate the true sugar content. Sample results: Sample size: 76 Sample Mean: 24.7 g Sample standard deviation: 1.3 g. Sample Mean: ..24.79 a. Provide the 90% Confidence Interval for the Mean amount of sugar per serving. b. The sample result of x = 24.7 g can be used as a point estimate for the mean amount of sugar per serving. At the 90% Confidence Level, what is the Margin of Error associated with this estimate?

Answer 1

(a) solution: Given that Sample size: 76 Sample mean: 24.79 Sample standard deviation : 1.39 901. Confidence soteeral ps the mean amount og sogar pese serving is [X + tn-1,/SIG) - 1-0 +1189 a=0.10 to-1,2)2 = 475, 0.10/2 = 1.6645 from to table) - 104.9.16645 x 13 | 176 ) = 0.90 = [04.9+0.2480] -0.90 =[ 24.6518, as. 148a ] is the required 90% confidence Intern

by Margin of Ellos M. E= thi, dla sirn 51.6645x1:31176 ME = 0.9482 of suos anociated with is 0.948. point estimate f = 24.7 i Maloin