Question

Suppose that a random sample of size 2 is to be selected, but this time sampling will be done with replacement. Using a method similar to that of Part (a), construct the sampling distribution of x. Hint: There are 16 different possible samples in this case. (Enter the probabilities as fractions.)

Answer 1

Answer:

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Step-by-step explanation:

The images for the complete question is attached to this solution.

The set S for the variables contains {5, 8, 11, 14}

For the (a) part, the set of possible samples of 2 without replacement include

(5,8), (5,11), (5,14), (8,5), (8,11), (8,14), (11,5), (11,8), (11,14), (14,5), (14,8), (14,11)

Note that it is impossible to pick a variable twice because the sampling is done without replacement.

- There are 12 different possible samples in this case.

Mean = x = μₛₐ = (x₁ + x₂)/2

The sample means of the sampling distribution include

(6.5), (8.0), (9.5), (6.5), (9.5), (11.0), (8.0), (9.5), (12.5), (9.5), (11.0), (12.5)

Sample means = x = μₛₐ; frequency = f

The sampling distribution of the means is now expressed in the table below.

The probabilities, p, are obtained from the relative frequency of each sample mean.

p = (f/Σf)

Σf = 12

μₛₐ | f | p

6.5 | 2 | (1/6)

8.0 | 2 | (1/6)

9.5 | 4 | (1/3)

11.0 | 2 | (1/6)

12.5| 2 | (1/6)

b) For random sample of 2 with replacements, the possible samples include

(5,5), (5,8), (5,11), (5,14), (8,5), (8,8), (8,11), (8,14), (11,5), (11,8), (11,11), (11,14), (14,5), (14,8), (14,11), (14,14)

- Note that it is possible to pick a variable twice because the sampling is done with replacement.

- There are truly 16 different possible samples in this case.

Mean = x = μₛₐ = (x₁ + x₂)/2

The sample means of the sampling distribution include

(5.0), (6.5), (8.0), (9.5), (6.5), (8.0), (9.5), (11.0), (8.0), (9.5), (11.0), (12.5), (9.5), (11.0), (12.5), (14.0)

Sample means = x = μₛₐ; frequency = f

The sampling distribution of the means is now expressed in the table below.

The probabilities, p, are obtained from the relative frequency of each sample mean.

p = (f/Σf)

Σf = 16

μₛₐ | f | p

5.0 | 1 | (1/16)

6.5 | 2 | (1/8)

8.0 | 3 | (3/16)

9.5 | 4 | (1/4)

11.0 | 3 | (3/16)

12.5| 2 | (1/8)

14.0| 1 | (1/16)

Hope this Helps!!!