Question

A

Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the night shift than on the day shift. A sample of 55 day-shift workers showed that the mean number of units produced was 342, with a population standard deviation of 26. A sample of 63 night-shift workers showed that the mean number of units produced was 349, with a population standard deviation of 32 units.

At the .10 significance level, is the number of units produced on the night shift larger?

(1)	This is a (Click to select)onetwo-tailed test.

(2)	The decision rule is to reject H0:μd≥μnH0: μd⁢≥μn if z < . (Negative value should be indicated by a minus sign. Round your answer to 2 decimal places.)

(3)	The test statistic is z = . (Negative value should be indicated by a minus sign. Round your answer to 2 decimal places.)

(4)	What is your decision regarding H0H0?
	(Click to select)Reject.Do not reject.

B.

A coffee manufacturer is interested in whether the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers. Assume the population standard deviation for those drinking regular coffee is 1.39 cups per day and 1.55 cups per day for those drinking decaffeinated coffee. A random sample of 45 regular-coffee drinkers showed a mean of 4.59 cups per day. A sample of 39 decaffeinated-coffee drinkers showed a mean of 5.19 cups per day.

Use the .05 significance level.

(1)	This is a (Click to select)onetwo-tailed test.

(2)	The decision rule is to reject H0 : μr ≥ μd if z < . (Negative value should be indicated by a minus sign. Round your answer to 2 decimal places.)

(3)	The test statistic is z = . (Negative value should be indicated by a minus sign. Round your answer to 2 decimal places.)

(4)	What is your decision regarding H0 ?
	(Click to select)Reject.Do not reject.

(5)	The p-value is . (Round your answer to 4 decimal places.)

Answer 1

A.

1) this is one tail test

2)

Level of Significance ,    α =    0.10

Z-critical value , Z* =        -1.282   [excel function =NORMSINV(α)]
The decision rule is to reject H0:μd≥μn if z <-1.28

3)

mean of sample 1,    x̅1=   342
population std dev of sample 1,   σ1 =    26
size of sample 1,    n1=   55
      
mean of sample 2,    x̅2=   349
population std dev of sample 2,   σ2 =    32
size of sample 2,    n2=   63
      
difference in sample means =    x̅1 - x̅2 =    -7
      
std error , SE =    √(σ1²/n1+σ2²/n2) =    5.3427
      
Z-statistic =    ((x̅1 - x̅2)-µd)/SE = -7 / 5.3427 = -1.31

4)

since, z stat < -1.28 , reject Ho

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B.

1)

1) this is one tail test

2)

Level of Significance ,    α =    0.05

Z-critical value , Z* =        -1.645   [excel function =NORMSINV(α)]

The decision rule is to reject H0:μr≥μd if z <-1.65

3)

mean of sample 1,    x̅1=   4.59
population std dev of sample 1,   σ1 =    1.39
size of sample 1,    n1=   45
      
mean of sample 2,    x̅2=   5.19
population std dev of sample 2,   σ2 =    1.55
size of sample 2,    n2=   39
      
difference in sample means =    x̅1 - x̅2 =    -0.6
      
std error , SE =    √(σ1²/n1+σ2²/n2) =    0.3233
      
Z-statistic =    ((x̅1 - x̅2)-µd)/SE = -0.6/0.3233 = -1.86

4)

since, z stat< critical value, reject Ho

5)

p-value =        0.0317   [excel function =NORMSDIST(z)]