Question

1. John has obtained two independent samples from two populations, where the sample statistics are shown in the table below. Assuming equal variances, he can construct a 95 percent confidence interval for the difference of the population means to be

	Sample 1	Sample 2
Mean	22.7	20.5
Variance (s^2)	5.4	3.6
Observations (sample size)	9	9

1. [0.08, 4.32]
2. [1.17, 5.08]
3. [2.44,6.19]
4. [-0.09, 3.19]

2. A corporate analyst is testing whether mean inventory turnover has increased. Inventory turnover in six randomly chosen product distribution centers (PDCs) is shown in the table below. Using a pair t-test, we find the value of the t statistic is equal to:

	This year	Last Year	Difference (d)
PDC 1	5.1	4.1	1
PDC 2	3.9	2.9	1
PDC 3	4.8	2.8	2
PDC 4	3.4	3.4	0
PDC 5	4.6	2.6	2
PDC 6	7.7	4.7	3
			S.D. of d= 1.048
			Mean of d=-1.5

1. 3.798
2. 2.449
3. 1.225
4. 3.503

3. True or False? A new policy of “flex hours” is proposed. Random sampling showed that 40 of 50 female workers favored the change, while 25 of 50 male workers favored the change. It seems the approval rate among female workers are greater than males. By a right-tailed test with α=0.025 where H0: pie 1- pie 2≤ 0.1 and H1: pie1 -pie2 greater than 0.1, we can reject the null hypothesis.
4. Refer to the following partial one-factor ANOVA results from excel. Now, the F statistic is equal to:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F Statistic
Between Groups			210.2788
Within Groups	1483		74.15
Total	2113.833

1. 4.79
2. 3.56
3. 1.15
4. 2.84

5. Referring to the table above (question #, the sum of squares for between groups variation is
   1. 129.99                
   2. 630.83
   3. 1233.4
   4. We cannot tell from the given information
6. Also referring to the table from question #5, the degrees of freedom for between groups variation is:
   1. 3
   2. 4
   3. 5
   4. 6

Answer 1

Level of Significance ,    α =    0.05
      
mean of sample 1,    x̅1=   22.7000
variance,s² = 5.4

size of sample 1,    n1=   9
      
mean of sample 2,    x̅2=   20.5
variance of sample 2 ,s^2 = 3.6
size of sample 2,    n2=   9
      
difference in sample means =    x̅1-x̅2 =    2.2000

Degree of freedom, DF=   n1+n2-2 =    16  
t-critical value =    t α/2 =    2.1199   (excel formula =t.inv(α/2,df)
          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    2.1213  
          
std error , SE =    Sp*√(1/n1+1/n2) =    1.0000  
margin of error, E =    t*SE =    2.1199  
          
difference of means =    x̅1-x̅2 =    2.2000  
confidence interval is           
Interval Lower Limit=   (x̅1-x̅2) - E =    0.08
Interval Upper Limit=   (x̅1-x̅2) + E =    4.32

so, option a) is answer

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sample size ,    n =    6
      
mean of sample 1,    x̅1=   4.9167
      
mean of sample 2,    x̅2=   3.4167
      
mean of difference ,    D̅ =   1.5000
      
std dev of difference , Sd =        1.0488
      
std error , SE =    Sd / √n =    0.4282
      
t-statistic =    (D̅ - µd)/SE =    3.503

so, answer is option d)

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