Question

Consider the following hypotheses:

H₀: μ = 470
H_A: μ ≠ 470

The population is normally distributed with a population standard deviation of 44. (You may find it useful to reference the appropriate table: z table or t table)

a-1. Calculate the value of the test statistic with x−x− = 483 and n = 65. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.)
  
Test statistic = ?

a-2. What is the conclusion at the 10% significance level?
  

A) Do not reject H₀ since the p-value is greater than the significance level.

B) Do not reject H₀ since the p-value is less than the significance level.

C) Reject H₀ since the p-value is greater than the significance level.

D) Reject H₀ since the p-value is less than the significance level.

a-3. Interpret the results at αα = 0.10.

A) We cannot conclude that the population mean differs from 470.

B) We conclude that the population mean differs from 470.

C) We cannot conclude that the sample mean differs from 470.

D) We conclude that the sample mean differs from 470.

b-1. Calculate the value of the test statistic with x−x− = 438 and n = 65. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.)
  
Test statistic = ?

b-2. What is the conclusion at the 5% significance level?
  

A) Reject H₀ since the p-value is greater than the significance level.

B) Reject H₀ since the p-value is less than the significance level.

C) Do not reject H₀ since the p-value is greater than the significance level.

D) Do not reject H₀ since the p-value is less than the significance level.

b-3. Interpret the results at αα = 0.05.

A) We conclude that the population mean differs from 470.

B) We cannot conclude that the population mean differs from 470.

C) We conclude that the sample mean differs from 470.

D) We cannot conclude that the sample mean differs from 470.

Answer 1

This is a 2 tailed test. Given = 470, = 44

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a-1: = 483, n = 65

The test statistic is given by the equation:

a-2: The Decision Rule is Reject H0 if p value is <

The p value(2 tail) at Z = 2.38. First we calculate the left tailed probability at Z = 2.38 which is = 0.99134. Since Z is positive, we find the 2 tailed probability as 2 * (1 - 0.99134) = 0.0173

Since p value (0.0173) is < (0.1), therefore

Option D: Reject H0 Since p value is less than the significance level.

a-3: Option B: We conclude that the population mean differes from 470.

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b-1: = 438, n = 65

The test statistic is given by the equation:

b-2: The Decision Rule is Reject H0 if p value is <

The p value(2 tail) at Z = -5.86. First we calculate the left tailed probability at Z = -5.86 which is = 0. Since Z is negative, we find the 2 tailed probability as 2 * 0 = 0.00

Since p value (0.010) is < (0.05), therefore

Option B: Reject H0 Since p value is less than the significance level.

b-3: Option A: We conclude that the population mean differes from 470.

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