We are now going to use Excel

We are now going to use Excel to test a hypothesis based on onegiven sample. We would like to know if the average cholesterollevel of patients in intensive care is equal to 200 and for thatreason we collect cholesterol level of 20 random people fromvarious intensive care units.

Cholesterol level154,168,134,201,208,220,225,228,201,207,168,211,203,254,268,198,298,135,154,189We put our data in column A. In Cells A3-A22 we type our data andin cell A1 we type Cholesterol. First let’s input the sample sizen, that is how many elements in data do we have. In cell B24 wetype N and in cell A24 we type 20. Second, we find degrees offreedom 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 cellB26 we type Mean and in cell A26 we type =average(A3:A22). Nowlet’s do standard deviation. In cell B27 we type SD and in cell A27we 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). Wewill do a one sample t test and for that we need a t value. . Incell B29 we type t and in cell A29 we type =(A26-200)/A28. Now weevaluate the p value using Excel. In cell B31 we type TEST valueand in cell A31 we type 200 In cell B32 we type Mean Difference. Incell A32 we type =A26-A31 . In cell B33 we type Sig.(2-tailed). Incell A33 type =T.DIST.2T(A29,A25). Here T.DIST.2T means we aredoing 2 tailed one sample TTEST. The first parameter A29 means weare computing the p value based on our calculated t value and weare comparing it to our threshold significance level of 0.05. Thesecond variable represents degrees of freedom.

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

Exercise 2. What does the value in cell A33 tell you about thehypothesis. Do we reject the null hypothesis? Why? Put yourexplanation 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. Putthat value in cell A37.

Exercise 4. Compare the t value in cell A29 to the table t valuein cell A37. What is this comparison telling you about rejecting ornot rejecting the null hypothesis. Put your explanation in cellA38.


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. errorof mean) 0.126574036
P- value is given as 0.900607429
Since P – value is > 0.05, we fail to reject Nullhypothesis

I have solved in detail as below for clear understanding :

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