Question

When σ is unknown and the sample is of size n ≥ 30, there are two methods for computing confidence intervals for μ. (Notice that, When σ is unknown and the sample is of size n < 30, there is only one method for constructing a confidence interval for the mean by using the Student's t distribution with d.f. = n - 1.) 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 = 30, with sample mean x = 45.2 and sample standard deviation s = 5.3. (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? Yes. The respective intervals based on the t distribution are longer. No. The respective intervals based on the t distribution are shorter. Yes. The respective intervals based on the t distribution are shorter. No. The respective intervals based on the t distribution are longer. (d) Now consider a sample size of 50. 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 (f) 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? Yes. The respective intervals based on the t distribution are longer. Yes. The respective intervals based on the t distribution are shorter. No. The respective intervals based on the t distribution are shorter. No. The respective intervals based on the t distribution are longer. With increased sample size, do the two methods give respective confidence intervals that are more similar? As the sample size increases, the difference between the two methods is less pronounced. As the sample size increases, the difference between the two methods becomes greater. As the sample size increases, the difference between the two methods remains constant.

Answer 1

Given = 45.2, s = 5.3

The Confidence interval is given by ME, where ME = Critical value (z or t) * s / Sqrt(n)

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(a) n = 30, therefore degrees of freedom = 29

90% CI using students t method

tcritical for = 0.10, critical value = 1.699.

Therefore ME = 1.699 * [5.3 / sqrt(30)] = 1.64

Lower Limit = 45.2 - 1.64 = 43.56

Upper Limit = 45.2 + 1.64 = 46.84

95% CI using students t method

tcritical for = 0.05, critical value = 2.045.

Therefore ME = 2.045 * [5.3 / sqrt(30)] = 1.98

Lower Limit = 45.2 - 1.98 = 43.22

Upper Limit = 45.2 + 1.98 = 47.18

99% CI using students t method

tcritical for = 0.01, critical value = 2.756.

Therefore ME = 2.756 * [5.3 / sqrt(30)] = 2.67

Lower Limit = 45.2 - 2.67 = 42.53

Upper Limit = 45.2 + 2.67 = 47.87

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(b) n = 30

90% CI using Normal Distribution

tcritical for = 0.10, critical value = 1.645.

Therefore ME = 1.645 * [5.3 / sqrt(30)] = 1.59

Lower Limit = 45.2 - 1.59 = 43.61

Upper Limit = 45.2 + 1.59 = 46.79

95% CI using Normal Distribution

tcritical for = 0.05, critical value = 1.96.

Therefore ME = 1.96 * [5.3 / sqrt(30)] = 1.90

Lower Limit = 45.2 - 1.90 = 43.30

Upper Limit = 45.2 + 1.90 = 47.10

99% CI using Normal Distribution

tcritical for = 0.01, critical value = 2.576.

Therefore ME = 2.576 * [5.3 / sqrt(30)] = 2.49

Lower Limit = 45.2 -2.49 = 42.71

Upper Limit = 45.2 + 2.49 = 47.69

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(c) Yes, the respective intervals based on the t distribution are longer

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(d) n = 50, therefore degrees of freedom = 49

90% CI using students t method

tcritical for = 0.10, critical value = 1.677.

Therefore ME = 1.677 * [5.3 / sqrt(50)] = 1.26

Lower Limit = 45.2 - 1.26 = 43.94

Upper Limit = 45.2 + 1.26 = 46.46

95% CI using students t method

tcritical for = 0.05, critical value = 2.01.

Therefore ME = 2.01 * [5.3 / sqrt(50)] = 1.51

Lower Limit = 45.2 - 1.51 = 43.69

Upper Limit = 45.2 + 1.51 = 46.71

99% CI using students t method

tcritical for = 0.01, critical value = 2.68.

Therefore ME = 2.68 * [5.3 / sqrt(50)] = 2.01

Lower Limit = 45.2 - 2.01 = 43.19

Upper Limit = 45.2 + 2.01 = 47.21

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(e) n = 50

90% CI using Normal Distribution

tcritical for = 0.10, critical value = 1.645.

Therefore ME = 1.645 * [5.3 / sqrt(50)] = 1.23

Lower Limit = 45.2 - 1.23 = 43.97

Upper Limit = 45.2 + 1.23 = 46.43

95% CI using Normal Distribution

tcritical for = 0.05, critical value = 1.96.

Therefore ME = 1.96 * [5.3 / sqrt(50)] = 1.47

Lower Limit = 45.2 - 1.47 = 43.73

Upper Limit = 45.2 + 1.47 = 46.67

99% CI using Normal Distribution

tcritical for = 0.01, critical value = 2.576.

Therefore ME = 2.576 * [5.3 / sqrt(50)] = 1.93

Lower Limit = 45.2 - 1.93 = 43.27

Upper Limit = 45.2 + 1.93 = 47.13

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(f) Yes, the respective intervals based on the t distribution are longer

(g) As the sample size increases, the difference between the 2 methods is less pronounced

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