Question 811546: 5. The average salary of senior managers in the construction industry is $180,000 per year. Suppose we would like to take a sample of senior managers at a newly established company XYZ to see whether the mean annual salary is different from that of the industry. (5 marks)
This is a two tailed test, because what I want to calculate is whether it is equal to, or not equal to x amount. I would call the industry mean o1 and company XYZ o2.
a. State the null and alternative hypotheses.
Null: Ho1 = Ho2: This states that the mean of one is the same as the second.
Alternative: Ho1 ≠ H02: This states that the mean of one is not the same as the second.
b. Suppose a sample of 40 senior managers at XYZ showed a sample mean annual salary of $170,000. Assume a population standard deviation of $30,000. With = .05 as the value of significance, what is your conclusion?
First, I calculate the test statistic by:
Sample mean-population mean/ standard dev. x square root of sample
So, it becomes 170,000-180,000/ 30,000 x 6.32
I calculate this to be: -0.0527
Because z< 0, this represents the area in the lower tail
This means that the area below -0.0527 shows a probability of 0.4801
Because it is a two tailed test, I would double this and calculate 0.9602.
Using the P-value approach, and the rule of (reject null is p < a), it says I should accept null which is incorrect as the means are obviously different.
The two tailed rejection rule is to reject null is z <-1.96 or z > 1.96
ic
Using the critical value approach, I can determine that my test statistic is less than 1.96, which is the zvalue for a/2.
Why is it right one way, and not the other?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The average salary of senior managers in the construction industry is $180,000 per year. Suppose we would like to take a sample of senior managers at a newly established company XYZ to see whether the mean annual salary is different from that of the industry. (5 marks)
This is a two tailed test, because what I want to calculate is whether it is equal to, or not equal to 180,000.
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a. State the null and alternative hypotheses.
Null: x-bar = 180,000.
Alternative: x-bar # 180,000
b. Suppose a sample of 40 senior managers at XYZ showed a sample mean annual salary of $170,000. Assume a population standard deviation of $30,000. With alpha = .05 as the value of significance, what is your conclusion?
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z(170,000) = (170,000-180,000)/[30,000/sqrt(40)] = -2/3
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p-value = 2*P(z < -2/3) = 2*normalcdf(-100,-2/3) = 0.5050
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Conclusion:: Since the p-value is greater than 5%, fail
to reject Ho.
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Cheers,
Stan H.
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