Given the following i.i.d. observations of a of random variable Y: 16.0, 12.0, 1

Question

Given the following i.i.d. observations of a of random variable Y: 16.0, 12.0, 14.0, 13.4, 16.0, 12.5 (a) Calculate the sample mean, sample variance, and sample standard deviation. (b) Calculate an alpha = 0.05 confidence interval for the mean, using the Central Limit Theorem approximation. (c) Calculate an alpha = 0.05 confidence interval for the mean, using Student's t-distribution. (A t-table is posted on the eLearning website.) (d) Estimate how many measurements you would need to make in order to get a confidence interval for the mean that is plusminus 2.0 wide. Answer this using both the Central Limit Theorem and the t-distribution.

Explanation / Answer

Part a

The calculation table for finding the sample mean, sample variance and sample standard deviation is given as below:

No.

Y

(Y - Ybar)

(Y - Ybar)^2

1

16

2.0167

4.06707889

2

12

-1.9833

3.93347889

3

14

0.0167

0.00027889

4

13.4

-0.5833

0.34023889

5

16

2.0167

4.06707889

6

12.5

-1.4833

2.20017889

Total

83.9

14.60833334

Mean

13.98333

Variance

2.921666668

SD

1.709288351

Mean = Y/n = 83.9/6 = 13.98333

Var = (Y – Ybar)^2/(n – 1) = 14.60833334/(6 – 1) = 2.9217

SD = sqrt(Var) = sqrt(2.9217) = 1.7093

Part b

Here, we have to use the z critical value for finding the confidence interval for the mean. Here, we have to use central limit theorem approach which suggests that the sampling distribution of the sample statistics approaches to the approximate normal distribution as we increase the sample size.

We are given alpha = 0.05.

Critical value z = 1.96

Confidence interval = Ybar -/+ z*SD/sqrt(n)

Confidence interval = 13.9833 -/+ 1.96*1.7093/sqrt(6)

Lower limit = 13.9833 – 1.3677 = 12.62

Upper limit = 13.9833 + 1.3677 = 15.35

Part c

Here, we have to calculate the same confidence interval by using t distribution.

We are given alpha = 0.05.

Sample size = n = 6

So, degrees of freedom = n - 1 = 5

Critical value t = 2.5706

Confidence interval = Ybar -/+ t*SD/sqrt(n)

Confidence interval = 13.9833 -/+ 2.5706*1.7093/sqrt(6)

Lower limit = 13.9833 – 1.7938 = 12.19

Upper limit = 13.9833 + 1.7938 = 15.78

Part d

First of all we have to find sample size by using central theorem approach.

Sample size = n = (Z*/E)^2

We are given = 0.05, so c = 1 – 0.05 = 0.95

Z = 1.96

= 1.7093

E = 2

n = (1.96*1.7093/2)^2 = 2.8059

Required sample size = 3

By using t distribution,

Critical t value = 2.5706

n = (2.5706*1.7093/2)^2 =4.826648

Required sample size = 5

No.

Y

(Y - Ybar)

(Y - Ybar)^2

1

16

2.0167

4.06707889

2

12

-1.9833

3.93347889

3

14

0.0167

0.00027889

4

13.4

-0.5833

0.34023889

5

16

2.0167

4.06707889

6

12.5

-1.4833

2.20017889

Total

83.9

14.60833334

Mean

13.98333

Variance

2.921666668

SD

1.709288351

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