Answer to Question #315929 in Statistics and Probability for mavis

Question

7. In 1900, W. Y. Kevin published the article “The anticipated Error of a Mean”. In this pioneering paper, written under the pseudonym “Student,” Kevin introduced what later became known as Student’s t-distribution. Kevin used the following data set, which gives the additional sleep in hours obtained by a sample of 10 patients using laevohysocyamine hydrobromide.

1.9 0.8 1.1 0.1 −0.1

4.4 5.5 1.6 4.6 3.4

a. Obtain and interpret a 95% confidence interval for the additional sleep that would be obtained on average for all people using laevohysocyamine hydrobromide.

b. Was the drug effective in increasing sleep? Explain your answer.

Answer 1

a)

$∑x_i=1.9+0.8+...+3.4=23.3$

$∑(x_i-\mu)^2=(1.9-2.33)^2+(0.8-2.33)^2+...+(3.4-2.33)^2=36.081$

$\mu=\frac{1}{n}∑x_i=\frac{23.3}{10}=2.33$

$s^2=\frac{1}{n-1}∑(x_i-\mu)^2=\frac{36.0810}{10-1}=4.0090$

From the student t-table

t-tabulated=$t_{0.05,10-1}=t_{0.05,9}$ =2.262

$\mu±t\sqrt{\frac{s^2}{n}}=2.33±2.262\sqrt{\frac{4.0090}{10}}$

$=2.33±1.4322$

$=0.8978<\mu<3.7622$

b)

The drug was effective in increasing sleep because zero is not included in this interval as the lower limit is greater than 0.8978