# We would like to find how much money the average shopper spends at the mall per trip. We sample...

## Question:

We would like to find how much money the average shopper spends at the mall per trip. We sample 75 shoppers and we find that their spending averages $49.80. Assume the standard deviation of the population is$16.55, find a 90% confidence interval for the true mean of the spending.

What is the upper limit of the interval?

## Confidence interval

The confidence interval includes a lower and upper limit which is an estimate using a margin of error. The maximum likely distinction between the observed value and true value of the population parameter.

Given information

• Sample size: 75
• Sample mean: 49.80
• Population standard deviation: 16.55

The critical value is obtained from the standard normal table at a level of significance (0.10)

The critical value is 1.645

The 90% confidence interval is calculated as follow.

{eq}\begin{align*} P\left( {\bar X - {Z_{\alpha /2}}\dfrac{\sigma }{{\sqrt n }} < \mu < \bar X + {Z_{\alpha /2}}\dfrac{\sigma }{{\sqrt n }}} \right) &= 0.90\\ P\left( {49.80 - 1.645\dfrac{{16.55}}{{\sqrt {75} }} < \mu < 49.80 + 1.645\dfrac{{16.55}}{{\sqrt {75} }}} \right) &= 0.90\\ P\left( {46.656 < \mu < 52.943} \right) &= 0.90 \end{align*}{/eq}

Therefore, required confidence interval is (46.656 to 52.943)

The upper limit of the interval is 52.943