Question

Use the data below to test whether the average deposits of customers have increased since the change. Use 0.05 as the level of significance. Note that the difference in Deposits is already calculated as Increase in Deposits.

Deposit After 30.6 18.1 19.3 31.0 21.9 21.3 24.1 18.4 19.6 18.9 30.6 19.3 29.0 21.7 18.6 20.4 27.6 26.7 27.7 19.8 19.3 20.1 18.2 18.5 25.3 30.8 30.1 23.6 30.1 23.8 25.3 26.0 25.4 31.9 26.2 29.6 19.1 25.6 23.0 18.9 21.9 25.3 25.9 30.0 20.7 30.4 31.6 28.5 20.6 20.6 27.6 30.0 27.8 22.2 20.4 19.2 21.2 24.0 19.0 22.2 31.7 27.5 19.1 21.3 20.7 20.3 29.8 31.6 26.6 25.8 27.9 18.5 27.7 22.2 20.1 28.9 18.4 28.9 21.3 22.5 31.3 22.3 20.4 25.9 23.9 21.3 23.2 22.2 18.7 19.3 28.5 22.6 22.9 26.4 29.4 21.7 19.9 19.5 27.4 28.9 31.3 25.3 27.1 22.9 29.6 25.8 28.7 26.9 29.3 23.1 20.5 18.0 18.6 23.7 25.9 29.2 28.6 22.8 27.7 27.0 25.1 25.5 25.8 25.6 30.2 31.7 26.2 30.2 31.2 30.1 21.9 28.2 27.1 26.5 21.0 27.2 26.3 29.2 26.4 22.6 18.6 25.8 18.6 27.4 32.0 25.6 30.6 18.3 18.8 18.8 18.8 22.0

Increase in Deposits 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

Answer 1

Hello,

the above problem is the example of the t-test.

Here Increase in Deposits is calculated. using the one sample t test we can solve the above problem.

A single sample t-test (or one sample t-test) is used to compare the mean of a single sample of scores to a known or hypothetical population mean.

Null Hypothesis

H₀: d = 0, where M is the sample mean and μ is the population or hypothesized mean.

h1: d > 0, ie Increase in Deposits

As above, the null hypothesis is that there is no difference between the sample mean and the known or hypothesized population mean.

Equation

d is means of Increase in Deposits, s is standard deviation of d.

Calculate the standard deviation in excel or calculator,

in excel input data and use command as =STDEV.S(A1:A152)

s= 5.8514

d=3.92

n=152

t=12.32

p-value in excel is = =T.DIST.RT(12.31,151) = >0.000

here p-value is less than the 0.05, so we reject the null hypothesis and conclude that the average deposits of customers have increased since the change.

Thanks