Question

IN EXCELL

We are now going to use Excel to test a hypothesis based on one given sample. We would like to know if the average cholesterol level of patients in intensive care is equal to 200 and for that reason we collect cholesterol level of 20 random people from various intensive care units. I Cholesterol level 154,168,134,201,208,220,225,228,201,207,168,211,203,254,268,198,298,135,154,189 We put our data in column A. In Cells A3-A22 we type our data and in cell Al we type Cholesterol. First let's input the sample size n, that is how many elements in data do we have. In cell B24 we type N and in cell A24 we type 20. Second, we find degrees of freedom df=N-1. In cell B25 we type df and in cell A25 we type =A24-1. Next we evaluate the mean. We learned that already. In cell B26 we type Mean and in cell A26 we type =average(A3:222). Now let's do standard deviation. In cell B27 we type SD and in cell A27 we type =stdey[13:A22). Next let's do standard error of the mean. In cell B28 we type SEM and in cell A28 we type =A27/sqrt(A24). We will do a one sample t test and for that we need at value. In cell B29 we type t and in cell A29 we type ={A26-200/A28. Now we evaluate the p value using Excel. In cell B31 we type TEST value and in cell A31 we type 200 in cell B32 we type Mean Difference. In cell A32 we type =A26-A31 In cell B33 we type Sig.(2-tailed). In cell A33 type =T.DIST.2T(A29,425). Here T.DIST.2T means we are doing 2 tailed one sample TTEST. The first parameter A29 means we are computing the p value based on our calculated t value and we are

Answer 1

A B I n D E F G - J K L 10 228 11 201 207 12 13 168 14 211 203 15 16 254 17 268 198 18 19 298 20 135 21 154 22 189 23 24 20 N 25 19 df 26 201.2 Mean 27 42.39861 SD 28 9.480617 SEM 29 0.126574 t 30 31 200 TEST 32 1.2 Mean Difference 33 0.900607 Sig (2-tailed) 34 35 Ho: u=200 36 We fail to reject the null hypothesis, as the the value in the cell A33 represents the p-value and it is more thwn the significance level. 37 The critical value at a significance level of 0.05 is 2.09 38 The obtained test statistic is less than the critical value, thus we fail to reject the null hypothesis.