Question

1. A student carries out an analysis of carbohydrate 5 times and finds the following percentages: 12.6, 11.9, 12.1, 10.5, and 13.1. Find the average and associated error at the 95% confidence interval

Answer 1

It is given that; A student carries out an analysis of carbohydrate 5 times and the data in the percentages is given below: 12.6 11.9 12.1 10.5 13.1 The average of the data is calculated as follows: na1 12.6+11.9+12.1+10.5 47.1 5 =9.42 Hence, the average of the data is = 9.42].

The sample standard deviation of the data is calculated as follows: 12.6-9.42)+(11.9-9.42) +(12.1-9.42) +(10.5-9.42) +(13.1-9.42) 5-1 38.154 = 9.5385 = 3.0884 Hence, the sample standard deviation of the data is s=3.0884 The formula for calculating the margin of error is given below: M.E = /2n-1) To The level of significance is a = 0.05. The degrees of freedom is calculated as follows: df =n-1 = 5-1 =4 Hence, the degrees of freedom is df = 4.

The critical value of t-statistic at 5% level of significance at 4 degrees of freedom is 10.05/40r = +2.776 The margin of error is calculated as follows: ME = +2.776x3.0884 8.5733984 = 23.8341 Hence, the margin of error is M.E = +3.8341 . The 95% confidence interval is calculated as follows: C.I =(+M.E) =(9.42+3.8341) = (5.5859,13.2541) Hence, the 95% confidence interval is (5.5859,13.2541).