# 5.4.3 - Example

5.4.3 - Example## Example 5-6: Emergency Room Wait Time

You are interested in the average emergency room (ER) wait time at your local hospital. You take a random sample of 50 patients who visit the ER over the past week. From this sample, the mean wait time was 30 minutes and the standard deviation was 20 minutes. Find a 95% confidence interval for the average ER wait time for the hospital.

#### Answer

Note that, \(t_{0.025,49}\approx z_{0.025}\) as the degrees of freedom is 49

- In Minitab choose
`Stat`>`Basic Statistics`>`1-Sample t`. - From the drop down box select the
`Summarized data option`button. (If you have the raw data you would use the default drop down of One or more samples, each in a column.) - Enter the sample size, sample mean, and sample standard deviation in their respective text boxes.
- Click the
`Options`button. The default confidence level is 95. If your desire another confidence level edit appropriately. - Click
`OK`and`OK`again.

#### Using Minitab: Emergency Room Wait Time Example

Referring to our prior example of average emergency room wait time from our discussion on confidence intervals for a population mean, our by-hand calculations produced a 95% confidence interval of 24.28 to 35.72 minutes. Recall the following for that example: sample size 50, sample mean 30, and sample standard deviation 20.

In Minitab following the above steps, we get a 95% confidence interval:

N | Mean | StDev | SE Mean | 95% CI |
---|---|---|---|---|

50 | 30.00 | 20.00 | 2.84 | (24.32, 35.68) |

The slight discrepancy between the estimates is due to our by-hand calculation using the t-value associated with 40 degrees of freedom since the table did not include a d.f. of 49. Minitab used a t-value for the actually 49 degrees of freedom. With the larger degrees of freedom comes a smaller t-value. This would result in a smaller margin of error and a narrower interval - precisely what we have here.

The mean length of certain construction lumber is supposed to be 8.5 feet. A random sample of 81 pieces of such lumber gives a sample mean of 8.3 feet and a sample standard deviation of 1.2 feet.

**Step 1: Check the conditions**: The sample size is large ($n\ge 30$), so we may continue using the value from the t-distribution as our multiplier.**Step 2: Construct the CI**: The degrees of freedom are $n-1=80$. If we use the table, with d.f of 60, $t_{0.025}=2$.The 95% confidence interval is \begin{align} &=\bar{x}\pm t_{0.025}\dfrac{s}{\sqrt{n}}\\ &=8.3\pm 2\dfrac{1.2}{\sqrt{81}}\\ &=8.3\pm 0.2667\\ &=(8.0333, 8.5667) \end{align}

**Step 1: Check the conditions**: The sample size is large ($n\ge 30$), so we may continue using the value from the t-distribution as our multiplier.**Step 2: Construct the CI**: The degrees of freedom are $n-1=80$. If we use the table, with d.f of 60, $t_{0.005}=2.66$. The 99% confidence interval is \begin{align} &=\bar{x}\pm t_{0.005}\dfrac{s}{\sqrt{n}}\\ &=8.3\pm 2.66\dfrac{1.2}{\sqrt{81}}\\ &=8.3\pm 0.3547\\ &=(7.9453, 8.6547) \end{align}

Reflecting back on interpretation of a proportion interval, we see the same basic structure: level of confidence, parameter of interest, lower and upper bounds.