9. Test whether the sample evidence indicates that the average time an employee

Question

9. Test whether the sample evidence indicates that the average time an employee stays with a company in their current positions is more than 3 years when a random sample of 121 employees yielded a mean of 3.15 years and s = 0.9. Use a = 0.05. Assume normal distribution.

10. An experiment was performed on a certain metal to determine if the strength is a function of heating time (hours). Results based on 30 metal sheets are given below. Use the simple linear regression model.
X = 60
X2 = 200
Y = 90
Y2 = 550
XY = 316

Find the estimated y intercept and slope and write the equation of the least squares regression line. Estimate Y when X is equal to 3 hours. Also determine the standard error, the Mean Square Error, the coefficient of determination and the coefficient of correlation and show their relation.

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Explanation / Answer

Solution:-

9)

H0: £ 3
HA: > 3

Since n is very large we can use Z test even when population standard deviation is not given or we can use the theoretically correct t- test.

Using the Z test the critical value for one tailed 5% level is 1.645. Using the t-test the critical value for df = 120 for one-tailed test is 1.658.

The calculated test statistic is : (X - mu)/(s/sqrtn) = (3.15-3)/(0.9/sqrt121)    = 1.833

Since the absolute value of calculated test statistic (1.833) is greater than the critical value we reject the Null at 5% level test. This is true for both Z and t test

10) We have, X = 60/30 = 2Y = 90/20 = 3
SSxx = X2 – (X)2/n = 200 - 602/30 = 80
SSyy = Y2 – (Y)2/n = 550 – 902/30 = 280
SSxy = XY – (XY)/n = 316 – 60*90/30 = 136
b1= SSxy/ SSxx = 136/80 = 1.7
b0 = Y - b1X = 3– 1.7*2 = - 0.4
Y = -0.4+ 1.7*X
(Strength) = - 0.4 + 1.7*Heating time
For heating time equal to 3 hours the predicted value of Strength (in some units) will be;
(Strength) Heating time = 3 = -0.4 + 1.7*3 = 4.7 units.
SSR = b1 SSxy = 1.7*136 = 231.2
SST is simply SSyy = 280
Coefficient of Determination R2 = SSR/SST = 231.2/280 = 0.8257 showing that the model explains 82.57% of variation in the dependent variable.
SSE = SST – SSR = 280 – 231.2 = 48.8
Mean Square Error (MSE) and is given by SSE/(n-2) = 48.8/ 28 = 1.743
The Standard Error of the Regression se = MSE = 1.743= 1.320.
Finally, the coefficient of correlation r = SSxy/( SSxx SSyy) = 136/( 80*280) = 0.9087Thus,

square r2= 0.90872= 0.8257 , the coefficient of determination calculated above.

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