Answer to Question #192965 in Statistics and Probability for ttttt

Question

To estimate the average number of customers entering and buying at the supermarket, the supervisor of that supermarket estimated the number of customers visiting every 5 minutes. She randomly selects 5-min intervals and counts the number of arrivals at the supermarket. The figure 58, 32, 41, 56, 80, 45, 29, 32, and 78 were obtained and tallied. The analysis assume that the number of arrivals is normally distributed. What is the 95% confidence interval to have an estimation of the mean value for all 5-min intervals?

Answer 1

Since the sample size is less than 30 and the population variance unknown, we use t distribution.

$\overline{x}=\frac{\sum x_i}{n}=\frac{58+32+41+56+80+45+29+32+78}{9}=50.11$

$s=\sqrt{\frac{\sum(x_i-\overline{x})^2}{n-1}}=\sqrt{\frac{(58-50.11)^2+(32-50.11)^2+...+(78-50.11)^2}{9-1}}=19.2966$

$95\%CI=\overline{x}\pm t_{\alpha/2}\cdot\frac{s}{\sqrt{n}}=50.11\pm2.306\cdot\frac{19.2966}{\sqrt{9}}=(35.277,64.943)$

We are 95% confident that the mean value for the number of customers arriving in 5 minute intervals is 35.277 and 64.943.