Question

Q1. A researcher is investigating a new financial awareness program, designed to reduce the amount of money people spend on credit cards. She randomly selects 20 people with credit cards to put through the program. She obtains the records of money spent in the year before the program, as well as the amount of money spent in the year after the program, and conducts the appropriate 2-sample t-test. What is the correct number of degrees of freedom for this test?

Q2. A shopper was interested in comparing prices between Woolworths and Coles. He found the cost of the same 12 items at 7 randomly selected Woolworths stores and 9 randomly selected Coles stores. Let   be the price at the i-th Woolworths store and   be the price at the i-th Coles store. The following data was obtained:

What is the pooled estimate of the standard deviation?

Q3. A manager implements a new efficiency program, designed to decrease the time it takes for works to process orders. A random sample of 10 workers are monitored and their average time take to process an order before and after the training is given below.

Person	1	2	3	4	5	6	7	8	9	10
Before (min)	40	41	21	34	35	27	25	42	37	16
After (min)	33	38	12	26	36	26	24	43	27	20

If the manager conducts the appropriate hypothesis test to assess whether this program has been successful or not (in decreasing the time taken), what is the relevant value of the test statistic? (define the difference as before - after)

Q4. This question continues Q3. What is the appropriate p-value for the hypothesis test in Q4? It might help to know that

p(T > 0.81) = p(T < -0.81) = 0.22

p(T > 2.14) =  p(T < -2.14) = 0.03

where T is the appropriate t-distribution for the hypothesis test in Q4.

Answer 1

1)

n=20

Degree of freedom, DF=   n - 1 =    19

...............

2)

data?

.........

3)

Sample #1	Sample #2	difference , Di =sample1-sample2	(Di - Dbar)²
40	33	7.00	13.69
41	38	3.00	0.09
21	12	9.00	32.49
34	26	8.00	22.09
35	36	-1.00	18.49
27	26	1.00	5.29
25	24	1.00	5.29
42	43	-1.00	18.49
37	27	10.00	44.89
16	20	-4.00	53.29

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	318	285	33.000	214.100

sample size ,    n =    10              
                      
mean of sample 1,    x̅1=   31.800              
                      
mean of sample 2,    x̅2=   28.500              
                      
mean of difference ,    D̅ =ΣDi / n =   3.300              
                      
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    4.8774              
                      
std error , SE = Sd / √n =    4.8774   / √   10   =   1.5424  

t-statistic = (D̅ - µd)/SE = (   3.3   -   0   ) /    1.5424   =   2.140

4)

Degree of freedom, DF=   n - 1 =    9

p-value =        0.03

....................

THANKS

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