When is unknown and the sample is of size n 30, there are two methods for comput

Question

When is unknown and the sample is of size n 30, there are two methods for computing confidence intervals for .

Method 1: Use the Student's t distribution with d.f. = n 1.
This is the method used in the text. It is widely employed in statistical studies. Also, most statistical software packages use this method.

Method 2: When n 30, use the sample standard deviation s as an estimate for , and then use the standard normal distribution.
This method is based on the fact that for large samples, s is a fairly good approximation for . Also, for large n, the critical values for the Student's t distribution approach those of the standard normal distribution.

Consider a random sample of size n = 36, with sample mean x = 45.9 and sample standard deviation s = 5.2.

(a) Compute 90%, 95%, and 99% confidence intervals for using Method 1 with a Student's t distribution. Round endpoints to two digits after the decimal.

(b) Compute 90%, 95%, and 99% confidence intervals for using Method 2 with the standard normal distribution. Use s as an estimate for . Round endpoints to two digits after the decimal.

c) Now consider a sample size of 71. Compute 90%, 95%, and 99% confidence intervals for using Method 1 with a Student's t distribution. Round endpoints to two digits after the decimal.

d) Compute 90%, 95%, and 99% confidence intervals for using Method 2 with the standard normal distribution. Use s as an estimate for . Round endpoints to two digits after the decimal.

Explanation / Answer

degree of freedom df =n-1

here as std error =std deviation/(n)1/2

and lower limit =sample mean -/+ t*std error

a)

std error =std deviation/(n)1/2 =0.8667

below is table of critical vlaue of t for 35 degree of freedom

hence

b)

here critical values of z are as follows:

therefore confidence interval:

c)

std error =0.6171

table for critical value of t at 70 degree of freedom:

hence

d)

for df=35 90% 95% 99% criitcal t 1.6896 2.03011 2.72381

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