Question #315929

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.


Expert's answer

a)

xi=1.9+0.8+...+3.4=23.3∑x_i=1.9+0.8+...+3.4=23.3

(xiμ)2=(1.92.33)2+(0.82.33)2+...+(3.42.33)2=36.081∑(x_i-\mu)^2=(1.9-2.33)^2+(0.8-2.33)^2+...+(3.4-2.33)^2=36.081

μ=1nxi=23.310=2.33\mu=\frac{1}{n}∑x_i=\frac{23.3}{10}=2.33

s2=1n1(xiμ)2=36.0810101=4.0090s^2=\frac{1}{n-1}∑(x_i-\mu)^2=\frac{36.0810}{10-1}=4.0090

From the student t-table

t-tabulated=t0.05,101=t0.05,9t_{0.05,10-1}=t_{0.05,9} =2.262


μ±ts2n=2.33±2.2624.009010\mu±t\sqrt{\frac{s^2}{n}}=2.33±2.262\sqrt{\frac{4.0090}{10}}

=2.33±1.4322=2.33±1.4322

=0.8978<μ<3.7622=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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

LATEST TUTORIALS
APPROVED BY CLIENTS