Question

1.Consider a population of values has a normal distribution with μ=193.1 and σ=89.5. You intend to draw a random sample of size n. The boxes are labeled #1-12. Complete the table by writing the number that goes in the box with the corresponding number. Round to four decimal places.

n	Sample means	Sample Standard Deviations
2	1.	2.
5	3.	4.
10	5.	6.
20	7.	8.
50	9.	10.
100	11.	12.

2. Describe what happens to the sample mean and sample standard deviation as your sample size increases.

3. If we were working with a sample size of 20, you would expect 95% of the samples to land in what interval around the sample means?

Fill in the blanks of the open interval and round to four decimal places (??)

4. If we were working with a sample size of 50, you would expect 95% of the samples to land in what interval around the sample means?

Fill in the blanks of the open interval and round to four decimal places (??)

5. What will happen to the previous intervals as the sample size increases?

6. When working with a sample size of 20, what percentage of the samples will be less than 200? Round to two decimal places.

7. Tell what calculator commands you used in the last question.

8. When working with a sample size of 20, what percentage of the samples will be greater than 190? Round to the 2nd decimal place.

Answer 1

Answer :-

Given that :-

• Sample mean ( μ or X̄ ) = 193.1
• Standard deviation ( ) = 89.5

1) A random sample of size (n)

2) If the sample mean and sample standard deviation as the sample size increase, then :-

=>The mean remains same, but standard deviation of the sampling distribution values are decrease. As the sample size increase.

3) We were working with a sample size = 20.

=> confidenc interval ( C.I ) = 95℅

=> n = 20

=> = 1 - confidence

= 1 - 0.95

=> = 0.05

( critical value Z = 1.96)

C.I = X̄ Z/2 /√n

= 193.1 1.96 (89.5/√20)

= 193.1 39.225

= [ 193.1 + 39.225, 193.1 - 39.225 ]

= C. I. = 232.325 , 153.875

4) We were working with a sample size = 50.

=> Confidence Interval (C. I) = 95℅

=> n = 50

( Critical value Z = 1.96)

C.I = X̄ Z/2 /√n

= 193.1 1.96 (89.5/√50)

= 193.1 24.808

= [ 193.1 + 24.808 , 193.1 - 24.808 ]

=> C.I = 217.908 , 168.292

5) The previous intervals as the sample size increase when -

As 'n' that is sample size increase and the width of Confidence Interval is becomes narrow.