You are interested in finding a 90% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 11 randomly selected physical therapy patients. Round answers to 3 decimal places where possible.

10,6,6,6,24,15,27,9,5,27,18
a. To compute the confidence interval use a distribution.

between

b. With 90% confidence the population mean number of visits per physical therapy patient is and visits.

c. If many groups of 11 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of visits per patient and about percent will not contain the true population mean number of visits per patient.

To find a 90% confidence interval for the mean number of visits for physical therapy patients based on the given data, we can follow these steps:

### Given Data:
\[
10, 6, 6, 6, 24, 15, 27, 9, 5, 27, 18
\]

### Step 1: Calculate the Sample Mean (\(\bar{x}\))
\[
\bar{x} = \frac{10 + 6 + 6 + 6 + 24 + 15 + 27 + 9 + 5 + 27 + 18}{11} = \frac{153}{11} \approx 13.909
\]

### Step 2: Calculate the Sample Standard Deviation (s)
\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
\]
First, calculate the squared deviations:
\[
(10 - 13.909)^2 = 15.255
\]
\[
(6 - 13.909)^2 = 62.462
\]
\[
(6 - 13.909)^2 = 62.462
\]
\[
(6 - 13.909)^2 = 62.462
\]
\[
(24 - 13.909)^2 = 101.836
\]
\[
(15 - 13.909)^2 = 1.188
\]
\[
(27 - 13.909)^2 = 171.564
\]
\[
(9 - 13.909)^2 = 24.091
\]
\[
(5 - 13.909)^2 = 79.991
\]
\[
(27 - 13.909)^2 = 171.564
\]
\[
(18 - 13.909)^2 = 16.689
\]

Sum of squared deviations:
\[
\sum (x_i - \bar{x})^2 = 769.564
\]
Now, calculate the sample standard deviation:
\[
s = \sqrt{\frac{769.564}{10}} \approx \sqrt{76.956} \approx 8.770
\]

### Step 3: Determine the t-Score for a 90% Confidence Interval
Since the sample size is small (n = 11), we use the t-distribution. Degrees of freedom (\(df\)) is \(n-1 = 10\). For a 90% confidence interval, the critical t-value (\(t_{\alpha/2}\)) is approximately 1.812.

### Step 4: Calculate the Margin of Error (ME)
\[
ME = t_{\alpha/2} \times \frac{s}{\sqrt{n}} = 1.812 \times \frac{8.770}{\sqrt{11}} \approx 4.792
\]

### Step 5: Compute the Confidence Interval
\[
\text{Lower bound} = \bar{x} - ME = 13.909 - 4.792 \approx 9.117
\]
\[
\text{Upper bound} = \bar{x} + ME = 13.909 + 4.792 \approx 18.701
\]

### Answers:
a. To compute the confidence interval, use the t-distribution.

b. With 90% confidence, the population mean number of visits per physical therapy patient is between **9.117** and **18.701** visits.

c. If many groups of 11 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About **90%** of these confidence intervals will contain the true population mean number of visits per patient, and about **10%** will not contain the true population mean number of visits per patient.

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Do you want more details or have any questions?

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### Related Questions:
1. How do you interpret a 90% confidence interval?
2. Why is the t-distribution used instead of the normal distribution in this case?
3. What would happen to the confidence interval if the sample size was larger?
4. How does the sample standard deviation affect the confidence interval?
5. What is the meaning of the margin of error in the context of this problem?
6. How would you calculate a 95% confidence interval for the same data?
7. What are the assumptions underlying the confidence interval calculation?
8. How can confidence intervals be used in decision-making?

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### Tip:
When working with small sample sizes (n < 30), it's important to use the t-distribution rather than the z-distribution, as it accounts for the additional uncertainty in estimating the population standard deviation.

Learn how to calculate a 90% confidence interval for the mean number of visits in physical therapy patients using the t-distribution. Includes step-by-step calculations with a detailed explanation of sample mean, sample standard deviation, and margin of error. Suitable for advanced high school students and beyond.

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