Find a 99 percent confidence interval for µ if = 98.6,s = 2,and n = 5.Assume that the sample is randomly selected from a normally distributed population.
A)[95.69 101.51]
B)[95.25 101.95]
C)[94.48 102.72]
D)[89.39 107.81]

Question

Find a 99 percent confidence interval for µ if  = 98.6,s = 2,and n = 5.Assume that the sample is randomly selected from a normally distributed population.
A)[95.69 101.51] 
B)[95.25 101.95] 
C)[94.48 102.72] 
D)[89.39 107.81]

Quizplus · Accepted Answer

We can use the t-distribution since the sample size is small (n=5). The formula for a 99% confidence interval for the population mean is:

x̄ ± tα/2 * s/√n

where x̄ is the sample mean, s is the sample standard deviation, n is the sample size, and tα/2 is the t-value with n-1 degrees of freedom and α/2 (0.005) level of significance.

From the information given, x̄ = 98.6, s = 2, n = 5. The t-value for α/2 = 0.005 and 4 degrees of freedom (n-1=4) is 4.604.

Plugging in the numbers, we get:

98.6 ± 4.604 * 2 / √5

= [94.48, 102.72]

Therefore, the best choice is C, which matches the calculated confidence interval.

Find a 99 Percent Confidence Interval for µ If