Question

Use the data in Bank Dataset to answer this question.

- Construct a 95% confidence interval for the mean increase in deposits. Note that the population standard deviation σ is not known in this case. Instead the sample standard deviation s should be calculated from the sample and the t distribution should be used.

2. What is the margin of error at the 95% confidence level?

Bank Dataset of Increase in deposits. Mean is 4. Sample size is 152 customers.

4.3

-6.3

1.9

7.7

4.4

5.7

-0.2

-1.6

-1.4

4.6

15.5

-5.5

3.4

0.1

-4.1

1.3

2.7

11.1

4.8

-3.3

0.1

5.3

-1.4

4.1

-1.0

7.6

5.8

2.9

6.7

-1.6

5.6

3.9

2.5

12.5

9.7

9.5

3.8

0.1

0.0

-2.7

2.8

5.0

1.3

15.7

-3.7

5.8

13.4

2.3

1.6

-6.0

7.6

11.0

8.1

6.0

-1.8

-7.4

-4.7

7.0

-7.9

-0.1

15.4

5.6

4.7

7.2

-2.4

-4.4

12.3

11.5

8.1

6.6

6.4

3.6

9.6

-4.1

-2.5

4.8

-4.0

11.8

6.5

2.8

13.8

8.2

3.2

7.6

6.4

4.6

-0.5

4.9

2.0

-6.5

8.7

-2.9

-3.0

5.3

8.5

-2.7

-3.0

-4.5

1.8

12.5

16.6

6.0

9.5

6.2

3.5

-0.7

11.4

4.9

13.1

0.7

5.7

2.8

-7.7

0.2

4.5

14.9

2.7

-2.8

11.0

1.1

8.1

2.4

0.4

1.0

11.3

10.3

0.7

15.6

9.1

10.8

5.6

10.2

10.2

3.3

-2.3

5.1

8.6

7.1

3.5

0.3

-4.6

11.5

4.5

12.1

9.8

5.9

14.9

-7.2

-5.0

-2.2

-2.5

-4.3

9.5

3.8

0.1

0.0

-2.7

2.8

5.0

1.3

15.7

-3.7

5.8

13.4

2.3

1.6

-6.0

7.6

11.0

8.1

6.0

-1.8

-7.4

-4.7

7.0

-7.9

-0.1

15.4

5.6

4.7

7.2

-2.4

-4.4

12.3

11.5

8.1

6.6

6.4

3.6

9.6

-4.1

-2.5

4.8

-4.0

11.8

6.5

2.8

13.8

8.2

3.2

7.6

6.4

4.6

-0.5

4.9

2.0

-6.5

8.7

-2.9

-3.0

5.3

8.5

-2.7

-3.0

-4.5

1.8

12.5

16.6

6.0

9.5

6.2

3.5

-0.7

11.4

4.9

13.1

0.7

5.7

2.8

-7.7

0.2

4.5

14.9

2.7

-2.8

11.0

1.1

8.1

2.4

0.4

1.0

11.3

10.3

0.7

15.6

9.1

10.8

5.6

10.2

10.2

3.3

-2.3

5.1

8.6

7.1

3.5

0.3

-4.6

11.5

4.5

12.1

9.8

5.9

14.9

-7.2

-5.0

-2.2

-2.5

-4.3

Answer #1

Here, n = 152

mean = 4

The sample standard deviation, as calculated with the help of Minitab software, is = 5.948

Now, degrees of freedom = (152 - 1) = 151.

At alpha = 0.05 and df = 151, the critical t value is 1.655

Therefore, 95% confidence interval = t*/n

= 4 1.655 * (5.948/21.236)

=( 4 - 0.801 , 4 + 0.801)

=( 3.199 , 4.801)

margin of error = t*/n

= 1.655 * (5.948/21.236)

= 0.801

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Build a stem-and-leaf plot for these measurements
3.1
4.9
2.8
3.6
2.5
4.5
3.5
3.7
4.1
4.9
2.9
2.1
3.5
4.0
3.7
2.7
4.0
4.4
3.7
4.2
3.8
6.2
2.5
2.9
2.8
5.1
1.8
5.6
2.2
3.4
2.5
3.6
5.1
4.8
1.6
3.6
6.1
4.7
3.9
3.9
4.3
5.7
3.7
4.6
4.0
5.6
4.9
4.2
3.1
3.9
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Year
Unemployment rate
gdp growth
cpi
inflation rate
1980
7.2
-0.2
87
1981
7.6
2.6
94
8.05
1982
9.7
-1.9
97.6
3.83
1983
9.6
4.6
101.3
3.79
1984
7.5
7.3
105.3
3.95
1985
7.2
4.2
109.3
3.80
1986
7.0
3.5
110.5
1.10
1987
6.2
3.5
115.4
4.43
1988
5.5
4.2
120.5
4.42
1989
5.3
3.7
126.1
4.65
1990
5.6
1.9
133.8
6.11
1991
6.9
-0.1
137.9
3.06
1992
7.5
3.6
141.9
2.90
1993
6.9
2.7
145.8
2.75
1994
6.1
4.0...

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0.8 0.9 1.2 1.3 1.5 1.6 1.8 1.8 2.3 2.4 2.8 2.8 2.9 2.9 3.3 3.8 4.4
4.6 5.9 6.5 6.8 7.2 7.3 7.8 8.8 9.3 9.6 11.1 11.4 11.4 12.9 13.1
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Use the following cell phone airport data speeds? (Mbps) from a
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0.1
0.2
0.2
0.4
0.4
0.5
0.5
0.6
0.6
0.7
0.7
0.7
0.8
0.8
0.8
0.9
1.1
1.6
1.7
1.8
1.8
2.1
2.4
2.5
2.7
2.9
3.4
3.5
3.5
3.9
4.6
4.6
5.7
6.5
7.2
7.5
8.9
10.3
10.5
10.8
11.3
11.4
12.3
12.3
12.9
13.3
14.2
14.8
15.9
28.6

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Calculate the R-chart and X-bar chart limits for the data given
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Day
A
B
C
D
1
7.2
8.4
7.9
4.9
2
5.6
8.7
3.3
4.2
3
5.5
7.3
3.2
6.0
4
4.4
8.0
5.4
7.4
5
9.7
4.6
4.8
5.8
6
8.3
8.9
9.1
6.2
7
4.7
6.6
5.3
5.8
8
8.8
5.5
8.4
6.9
9
5.7
4.7
4.1
4.6
10
3.7
4.0
3.0
5.2
11
2.6
3.9...

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1.3
5.8
6.1
5.6
7.3
7.1
4.7
2.8...

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0.7
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6.4
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