In: Statistics and Probability
A chemist made six independent measurements of the sublimation point of carbon dioxide (the temperature at which it changes to dry ice). She obtained a sample mean of 196.64 K with a standard deviation of 0.67 K. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.
Use the Student’s t distribution to find a 95% confidence interval for the sublimation point of carbon dioxide. (Round the final answers to two decimal places.)
Use the Student’s t distribution to find a 98% confidence interval for the sublimation point of carbon dioxide. (Round the final answers to two decimal places.)
Solution :
Given that,
Point estimate = sample mean = = 196.64
sample standard deviation = s = 0.67
sample size = n = 6
Degrees of freedom = df = n - 1 = 6 - 1 = 5
a) At 95% confidence level
= 1 - 95%
=1 - 0.95 =0.05
/2
= 0.025
t/2,df
= t0.025,5 = 2.571
Margin of error = E = t/2,df * (s /n)
= 2.571 * ( 0.67/ 6)
Margin of error = E = 0.70
The 95% confidence interval estimate of the population mean is,
± E
= 196.64 ± 0.70
= ( 195.94, 197.34 )
b) At 98% confidence level
= 1 - 98%
=1 - 0.98 =0.02
/2
= 0.01
t/2,df
= t0.01,5 = 3.365
Margin of error = E = t/2,df * (s /n)
= 3.365 * ( 0.67/ 6)
Margin of error = E = 0.92
The 98% confidence interval estimate of the population mean is,
± E
= 196.64 ± 0.92
= ( 195.72, 197.56 )