Question

# Use the data in Bank Dataset to answer this question. Construct a 95% confidence interval for...

Use the data in Bank Dataset to answer this question.

1. 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

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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