Question

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

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 σ, 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 σ. 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 = 31, with sample mean x = 46.0 and sample standard deviation s = 6.0.

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

	90%	95%99%
lower limit
upper limit

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

	90%	95%99%
lower limit
upper limit

(c) Compare intervals for the two methods. Would you say that confidence intervals using a Student's t distribution are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribution?

No. The respective intervals based on the t distribution are longer.

Yes. The respective intervals based on the t distribution are shorter. Yes. The respective intervals based on the t distribution are longer.

No. The respective intervals based on the t distribution are shorter.

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

	90%	95%99%
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upper limit

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

	90%	95%99%
lower limit
upper limit

Answer 1

Answer

(a)

Since there 31 data values in the sample so degree of freedom is df=31-1=30 and critical value of t for 90% confidence interval is 1.697. Therefore required confidence interval is

Critical value of t for 95% confidence interval is 2.042. Therefore required confidence interval is

Critical value of t for 99% confidence interval is 2.75. Therefore required confidence interval is

(b)

Critical value of z for 90% confidence interval is 1.645. Therefore required confidence interval is

Critical value of z for 95% confidence interval is 1.96. Therefore required confidence interval is

Critical value of z for 99% confidence interval is 2.58. Therefore required confidence interval is

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Now n=81

(c)

Since there 81 data values in the sample so degree of freedom is df=81-1=80 and critical value of t for 90% confidence interval is 1.664. Therefore required confidence interval is

Critical value of t for 95% confidence interval is 1.99. Therefore required confidence interval is

Critical value of t for 99% confidence interval is 2.639. Therefore required confidence interval is

(d)

Critical value of z for 90% confidence interval is 1.645. Therefore required confidence interval is

Critical value of z for 95% confidence interval is 1.96. Therefore required confidence interval is

Critical value of z for 99% confidence interval is 2.58. Therefore required confidence interval is