Question

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.

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 A1 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:A22). Now let’s do standard deviation. In cell B27 we type SD and in cell A27 we type =stdev(A3: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 a t 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,A25). 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 comparing it to our threshold significance level of 0.05. The second variable represents degrees of freedom.

Exercises Exercise 1. State the null hypothesis from this example in cell A35.

Exercise 2. What does the value in cell A33 tell you about the hypothesis. Do we reject the null hypothesis? Why? Put your explanation in cell A36.

Exercise 3. Instead of using Excel, we can look at the t chart. Find the critical value for the t distribution from that table. Put that value in cell A37.

Exercise 4. Compare the t value in cell A29 to the table t value in cell A37. What is this comparison telling you about rejecting or not rejecting the null hypothesis. Put your explanation in cell A38.

Answer 1

Cholesterol
154
168
134
201
208
220
225
228
201
207
168
211
203
254
268
198
298
135
154
189

sample mean	201.2		N (sample size)	20
sample std. deviation	42.39861		Degree of freedom , df = N-1	19
Std. error of mean	9.480617
	Exer. 1	Null Hypothesis ; H0 :	mu = 200
		Hypothesized mean	200
	Exer.2	Test statistic (t = (sample mean - hypothesized mean)/(Std. error of mean)	0.126574036
		P- value is given as	0.900607429
		Since P - value is > 0.05, we fail to reject Null hypothesis

I have solved in detail as below for clear understanding :