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Cash withdrawals from a college credit union for a random sample of 30 Fridays a

ID: 3043561 • Letter: C

Question

Cash withdrawals from a college credit union for a random sample of 30 Fridays and 30 Mondays are shown. At = .05, is there a difference in the mean withdrawal on Monday and Friday? The data are shown below and may be found in the data filed named ATM. 23 points

a)First test to determine if the variances in cash withdrawals differ between Friday and Monday. Show and follow the 7 steps.  

b)Verify your results in part (a) using Minitab.

c)Using the information from your results in part (a), test to determine if the average cash withdrawals differ between Friday and Monday. Follow and show the 7 steps for hypothesis testing.

e)Verify your results in part (c) using Minitab.

Randomly Chosen Cash Withdrawals ($)

Friday

Monday

250

10

10

40

30

10

20

10

30

100

70

370

110

20

10

20

20

10

40

20

40

30

50

30

70

10

10

200

20

40

20

20

400

20

30

20

10

20

10

10

20

100

50

20

10

30

40

20

100

20

20

50

10

20

20

60

70

60

10

20

Randomly Chosen Cash Withdrawals ($)

Friday

Monday

250

10

10

40

30

10

20

10

30

100

70

370

110

20

10

20

20

10

40

20

40

30

50

30

70

10

10

200

20

40

20

20

400

20

30

20

10

20

10

10

20

100

50

20

10

30

40

20

100

20

20

50

10

20

20

60

70

60

10

20

Explanation / Answer

a)First test to determine if the variances in cash withdrawals differ between Friday and Monday. Show and follow the 7 steps.

Solution:

Here, we have to use two sample F test for population variances. The null and alternative hypotheses are given as below:

Null hypothesis: H0: 1^2 = 2^2

Alternative hypothesis: Ha: 1^2 2^2

(Two tailed test)

We are given

Level of significance = = 0.05

Test statistic is given as below:

F = S1^2/S2^2

From given data, we have

Sample sizes

n1 = 30, n2 = 30

S1^2 = 6672.298851

S2^2 = 5131.034483

F = 6672.298851/ 5131.034483

F = 6672.298851/ 5131.034483 = 1.3004

Upper critical value = 2.1010

P-value = 0.4338

P-value > Level of significance =

So, we do not reject the null hypothesis.

Variances of cash withdrawals at Friday and Monday are same.

b)Verify your results in part (a) using Minitab.

Required Minitab results are given as below:

Test for Equal Variances

Level1     Friday

Level2     Monday

ConfLvl    95.0000

Bonferroni confidence intervals for standard deviations

Lower     Sigma     Upper     N Factor Levels

63.0506   81.6841   114.893    30 Friday

55.2909   71.6312   100.753    30 Monday

F-Test (normal distribution)

Test Statistic: 1.300

P-Value       : 0.484

Levene's Test (any continuous distribution)

Test Statistic: 0.038

P-Value       : 0.847

Results in part a and b are same.

c)Using the information from your results in part (a), test to determine if the average cash withdrawals differ between Friday and Monday. Follow and show the 7 steps for hypothesis testing.

Solution:

Here, we have to use two sample t test for population means.

H0: µ1 = µ2

Ha: µ1 µ2

(Two tailed test)

Test statistic formula is given as below:

t = (X1bar – X2bar) / sqrt[Sp2*((1/n1)+(1/n2))]

Where Sp2 is pooled variance

Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

From given data, we have

Level of significance = alpha = 0.05

X1bar = 50.33

S1 = 81.68

N1 = 30

X2bar = 50

S2 = 71.63

N2 = 30

Total degrees of freedom = N1 + N2 – 2 = 30 + 30 – 2 = 58

Critical values = -2.0017 and 2.0017

Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

Sp2 = [(30 – 1)*81.68^2 + (30 – 1)*71.63^2]/(30 + 30 – 2)

Sp2 = 5901.6667

t = (X1bar – X2bar) / sqrt[Sp2*((1/n1)+(1/n2))]

t = (50.33 – 50) / sqrt[5901.6667*((1/30) + (1/30))]

t = 0.0168

P-value = 0.9866

P-value > Alpha value

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that average cash withdrawals differ between Friday and Monday.

e)Verify your results in part (c) using Minitab.

Required Minitab results are given as below:

Two-Sample T-Test and CI: Friday, Monday

Two-sample T for Friday vs Monday

         N      Mean     StDev   SE Mean

Friday 30      50.3      81.7        15

Monday 30      50.0      71.6        13

Difference = mu Friday - mu Monday

Estimate for difference: 0.3

95% CI for difference: (-39.4, 40.0)

T-Test of difference = 0 (vs not =): T-Value = 0.02 P-Value = 0.987 DF = 58

Both use Pooled StDev = 76.8

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