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%
lower limit
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

n = 31. 2.-44.2. s 5.5

(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

끊 ± t 44.2 ± 1.697 44.2 ± 1.676 (42.5, 45.9) = critical ·- V31

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

5.5 it t 44.2 ± 2.042 . = 44.2 ±2.017 = (42.2, 46.2) critical v 31

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

5.5 44.2 ± 2. 75.-44.2+ 2.717 -(41.5, 46.9) critical Vn V31

(b)

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

5.5 ° 1= 44.2士 1.645· --= v/31 (42.6.45.8) T土~critical 44.2 ± 44.2± 1.625 Vn

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

5.5 --= V31 44.2 ± 1.96 44.2 ± 1.936 (42.3. 46.1) T E critical Vn

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

-± ~critical 442t 2.58 .--- 58 44.2 ± 2.549 (41.7, 46.7) = Vn

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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

5.5 ± t critical = 44.2 ± 1.664 44.2 1.017 (43.2, 45.2) V81

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

5.5 ± t critical 44.2 ± 1. 99- = 44.2± 1.216 = (43.0, 45.4) 44.2 +1.216 (43.0, 45.4) V81

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

55 ± tcriticarsn = 44.2± 2.639· . 639 44.2 ± 1.613 (42.6.45.8) = VSI 81

(d)

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

5.5 -criticalア Io = 44.2 ± 1.645- 44.2 ± 1.005 (43.2. 45.2) Vn 81

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

T E rt zcritcal 44.2 ± 1.96 --= 44.2 ± 1.2 (43.45.4) Vn V81

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

5.5 -± ~critical 44.2 ± 2.58 442t 2.58 .--- 44.2 ± 1.577 (42.6, 45.8) V81