A marketing consultant observed 50 consecutive shoppers at a supermarket. One variable of interest was how much each shopper spent in the store. Here are the data (in dollars), arranged in increasing order:

```
3.11, 8.88, 9.26, 10.81, 12.69, 13.78, 15.23, 15.62, 17, 17.39,
18.36, 18.43, 19.27, 19.5, 19.54, 20.16, 20.59, 22.22, 23.04, 24.47,
24.58, 25.13, 26.24, 26.26, 27.65, 28.06, 28.08, 28.38, 32.03, 34.98,
36.37, 38.64, 39.16, 41.02, 42.97, 44.08, 44.67, 45.4, 46.69, 48.65,
50.39, 52.75, 54.8, 59.07, 61.22, 70.32, 82.7, 85.76, 86.37, 93.34
```

Assume that the standard deviation is \$22.00. Find a 95% confidence interval for the mean amount spent by shoppers in similar circumstances.

### Solution

Given that sample size $n = 50$, sample mean $\overline{X}= 34.702$ and population standard deviation $\sigma = 22$.

#### Step 1 Specify the confidence level $(1-\alpha)$

Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.

#### Step 2 Given information

Given that sample size $n =50$, sample mean $\overline{X}=34.7022$ and population standard deviation is $\sigma = 22$.

#### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval for the population mean $\mu$ is

` $$ \begin{aligned} \overline{X} - E \leq \mu \leq \overline{X} + E \end{aligned} $$ `

where `$E = Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$`

, and `$Z_{\alpha/2}$`

is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

#### Step 4 Determine the critical value

The critical value of $Z$ for given level of significance is `$Z_{\alpha/2}$`

.

Thus `$Z_{\alpha/2} = Z_{0.025} = 1.96$`

.

#### Step 5 Compute the margin of error

The margin of error for mean is

` $$ \begin{aligned} E & = Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\\ & = 1.96 \frac{22}{\sqrt{50}} \\ & = 6.098. \end{aligned} $$ `

#### Step 6 Determine the confidence interval

$95$% confidence interval estimate for population mean is

` $$ \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 34.702 - 6.098 & \leq \mu \leq 34.702 + 6.098\\ 28.604 & \leq \mu \leq 40.8. \end{aligned} $$ `

Thus, $95$% confidence interval estimate for population mean is $(28.604,40.8)$.

#### Interpretation

We are $95$% confident that the true population mean is between $28.604$ and $40.8$.