Answer to Question #100230 in Statistics and Probability for sofia

Question

A research is conducted by a marketing research company in order to review the amount of customer’s spending before and after the promotion. The table below shows the spending of a sample of 12 customers in October (before the promotion) and in November (after the promotion):Customer 1 2 3 4 5 6 7 8 9 10 11 12
Spending in October
3500 4200 2675 2420 3450 2990 3430 3480 3720 4455 3830 3980
Spending in November
3920 4505 3147 2940 3786 3210 3620 3747 3810 4480 3970 4000
a. Use D to denote the amount of money a customer spends more in November than in October. Construct a 90% confidence interval estimate for the population mean of D.

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} Customer & \begin{array}{c} Spending \\ in \\ October \end{array} & \begin{array}{c} Spending \\ in \\ November \end{array} & D & \\ \hline 1 & 3500 & 3920 & 420 \\ \hline 2 & 4200 & 4505 & 305 \\ \hline 3 & 2675 & 3147 & 472 \\ \hline 4 & 2420 & 2940 & 520 \\ \hline 5 & 3450 & 3786 & 336 \\ \hline 6 & 2990 & 3210 & 220 \\ \hline 7 & 3430 & 3620 & 190 \\ \hline 8 & 3480 & 3747 & 267 \\ \hline 9 & 3720 & 3810 & 90 \\ \hline 10 & 4455 & 4480 & 25 \\ \hline 11 & 3830 & 3970 & 140 \\ \hdashline 12 & 3980 & 4000 & 20 \end{array}$$

$$N=12, \sum x_i=3005$$

$$mean=\bar{x}={\sum x_i \over N}={3005 \over 12}\approx250.4167$$

$$s^2={\sum (x_i-\bar{x})^2 \over N-1}\approx{307516.9167 \over 12-1}\approx27956.0833$$

$$s=\sqrt{s^2}\approx167.2007$$

Margin of Error

$$s_{\bar{x}}={s \over \sqrt{N}}$$

The critical value for $\alpha=0.1$ and $df=N-1=11$ degrees of freedom is $t_c=t_{1-\alpha/2;N-1}=1.796$

The corresponding confidence interval is computed as

$$CI=\bigg(\bar{x}-t_c\cdot{s \over \sqrt{N}}, \bar{x}+t_c\cdot{s \over \sqrt{N}}\bigg)$$

$$=(250.4167-1.796\cdot{167.2007 \over \sqrt{12}}, 250.4167+1.796\cdot{167.2007 \over \sqrt{12}})$$

$$=(163.7297,337.1037)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $163.7297<\mu<337.1037,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval $(163.7297,337.1037).$