Question

A Statistics Study is created by choosing for each question on the Study one possible version at random from a bank of possible versions of the question. There are 20 versions in the bank for each question. A specific question on the Study involves a one-sample test for the population mean with hypotheses

H₀ : µ = 15

H_a : µ > 15

with all versions of the question involving a sample of size n = 35.

use significance level (alpha) = 0.05

                                    alpha/2 = 0.025

Seven versions of the question give the population standard deviation as σ = 3.

Six versions give the sample standard deviation as s = 4.2.

The remaining versions give the sample standard deviation as s = 5.7.

Let c ∗ be the critical value for the rejection region on this question. Calculate E[c ∗ ].

Answer 1

Answer:

Given that:

A statistic Study is created by choosing for each question on the Study one possible version at random from a bank of possible versions of the question.

We assume that the significance level to be 0.05.

Critical value of z for 0.05 significance level is -1.645

Degree of freedom = n-1

= 35-1

= 34

Critical value of t for 0.05 significance leve and df = 34 is -1.691

For \sigma = 3

Standard error = 3 / \sqrt{35} = 0.5070926

Since we know the population standard deviation, we will use z score to estimate the critical value.

c = 15 - 1.645 0.5070926

= 14.16583

For s = 4.2

Standard error = 4.2 / \sqrt{35}

= 0.7099296

Since we do not know the population standard deviation, we will use t statistic to estimate the critical value.

c = 15 - 1.691 0.7099296

= 13.79951

For s = 4.2

Standard error = 5.7 / \sqrt{35}

= 0.9634759

Since we do not know the population standard deviation, we will use t statistic to estimate the critical value.

c = 15 - 1.691 0.9634759

= 13.37076

The PMF of C* is,

P(c* = 14.16583) = 7/20

P(c* = 13.79951) = 6/20

P(c* = 13.37076) = 1 - 7/20 - 6/20 = 7/20

E[c*] = (7/20) 14.16583 + (6/20) 13.79951 + (7/20) * 13.37076

= 13.77766