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A Statistics exam is created by choosing for each question on the exam 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 exam involves a one-sample test for the population mean with hypotheses:
H0 : µ = 15
Ha : µ < 15
with all versions of the question involving a sample of size n = 35.
Seven versions of the question give the population standard deviation as σ = 3. Six versions give the standard deviation as σ = 2. The remaining versions give the sample standard deviation as σ = 5.7.
Let c* be the critical value for the rejection region on this question. Calculate E[c*].

Respuesta :

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Answer: E (c *) = 13.77766

Step-by-step explanation:

significance level of 0.05

we will calculate the population mean with the following formula

H0: µ = 15

Ha: µ > 15

sample size n = 35

Degrees of freedom df = n-1 = 35-1 = 34

The critical value for z is -1.645

The critical value for t is -1.691

For \ sigma = 3,

Standard error = 3 / √ 35 = 0.5070926

the z-score is used to estimate the critical value since we know the standard deviation of the population

ex = 15-1645 * 0.5070926 = 14.16583

For s = 4.2,

Standard error = 4.2 / √ 35 = 0.7099296

next we will use the statistical t to estimate the critical value since we do not have the standard deviation of the population

c * = 15-1.691 * 0.7099296 = 13.79951

for s = 5.7,

Standard error = 5.7 /√ 3.5 = 0.9634759

c * = 15-1.691 * 0.9634759 = 13.37076

Let c * be the critical value for the rejection region in this question

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

finally the result:

E (c *) = 13.77766

Answer:

E[c*]=13.987

Step-by-step explanation:

We have different standard deviations for the test of hypothesis, as versions of the same problem.

The population mean is µ=15 and the sample size is n=35.

The critical value of z (zc) for an assumed level of confidence of 0.05 and a left-tail test is zc=-1.645.

Then, we can express the critical value c* as:

[tex]c^*=\mu+z\cdot \sigma/\sqrt{n}[/tex]

Its expected value can then be calculated as:

[tex]E(c^*)=E(\mu+z\cdot \sigma/\sqrt{n})=\mu+z\cdot\dfrac{E(\sigma)}{\sqrt{n}}[/tex]

The expected value of the standard deviation is:

[tex]E(\sigma)=\dfrac{1}{n}\sum_{i=1}^nn_i \sigma_i=\dfrac{1}{20}(7*3+6*2+7*5.7)\\\\\\\\E(\sigma)=\dfrac{1}{20}(21+12+39.9)=\dfrac{72.9}{20}=3.645[/tex]

Then, the expected value for the critical value is:

[tex]E(c^*)=\mu+z\cdot\dfrac{E(\sigma)}{\sqrt{n}}=15-1.645\cdot\dfrac{3.645}{\sqrt{35}}\\\\\\E(c^*)=15-1.645\cdot0.616=15-1.013=13.987[/tex]

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