Question

In 1908, W.S. Gosset published the article "The Probable Error of a Mean" (Biometrika, Vol. 6, pp. 1-25). In this pioneering paper, written under the pseudonym "Student," Gosset introduced what later became known as Student's t-distribution. Gosset used the following data set, which gives the additional sleep in hours obtained by a sample of 10 patients using laevohysocyamine hydrobromide (hereinafter referred to as "the drug").

1.9

0.8

1.1

0.1

-0.1

4.4

5.5

1.6

4.6

3.4

a. Create a 95% confidence interval for the additional sleep that would be obtained on average for all people using the drug. Briefly explain the process that you used and why.

b. Give an interpretation of the confidence interval that you found in part a. That is, what does it mean? c. Make an inference: Was the drug effective in increasing sleep? Explain your reasoning.

Reference

Weiss, N., (2008) Elementary statistics, (7th Ed.). Boston, MA: Pearson.

Answer 1

a)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   2.0022
Sample Size ,   n =    10
Sample Mean,    x̅ = ΣX/n =    2.3300

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   9          
't value='   tα/2=   2.2622   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   2.002   / √   10   =   0.6332
margin of error , E=t*SE =   2.2622   *   0.633   =   1.432
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    2.33   -   1.432   =   0.898
Interval Upper Limit = x̅ + E =    2.33   -   1.432   =   3.762
95%   confidence interval is (   0.90   < µ <   3.76   )

b)

we are 95% confident that true population mean of additional sleep lies within the confidence interval.

c)

Yes, the drug effective was in increasing sleep. because confidence interval is positive.