ANSWER

Statistical Analysis of Patient Numbers at Philip Sherman, DDS Dental Clinic

Introduction

In this analysis, we will examine the number of patients seen per day at Philip Sherman, DDS, dental clinic over a sample of 8 days in the past six months. Our goal is to estimate the variance of the number of patients seen per day and conduct a hypothesis test to determine if the variance is equal to 14.

Part A: Confidence Interval for Variance

To compute a 95% confidence interval estimate for the variance of the number of patients seen per day, we will follow these steps:

Step 1: Calculate the sample variance (s^2) of the given data.
Step 2: Determine the degrees of freedom (df) for the sample, which is the sample size (n) minus 1.
Step 3: Define the critical values of the chi-square distribution corresponding to the 0.025 and 0.975 probabilities for a 95% confidence interval.
Step 4: Compute the lower and upper bounds of the confidence interval.

Step 1: Sample Variance (s^2)
The sample mean (x̄) can be calculated by summing all the values and dividing by the number of days (n):

x̄ = (21 + 20 + 16 + 21 + 18 + 22 + 23 + 15) / 8 = 19.5

The sample variance (s^2) can be computed as follows:

s^2 = Σ (xi – x̄)^2 / (n – 1)
= [(21 – 19.5)^2 + (20 – 19.5)^2 + (16 – 19.5)^2 + (21 – 19.5)^2 + (18 – 19.5)^2 + (22 – 19.5)^2 + (23 – 19.5)^2 + (15 – 19.5)^2] / (8 – 1)
≈ 11.67

Step 2: Degrees of Freedom (df)
The degrees of freedom (df) for this sample is given by (n – 1):

df = 8 – 1 = 7

Step 3: Critical Values
For a 95% confidence interval, we need to find the critical values of the chi-square distribution with 7 degrees of freedom. From statistical tables or software, we find the critical values to be approximately 2.17 and 16.01.

Step 4: Confidence Interval
Now, we can calculate the lower and upper bounds of the confidence interval for the variance (σ^2):

Lower bound = (n – 1) * s^2 / critical value
= 7 * 11.67 / 16.01
≈ 5.11

Upper bound = (n – 1) * s^2 / critical value
= 7 * 11.67 / 2.17
≈ 38.03

The 95% confidence interval for the variance of the number of patients seen per day is approximately [5.11, 38.03].

Part B: Hypothesis Test for Variance

To conduct the hypothesis test, we set up the null and alternative hypotheses as follows:

Null hypothesis (H0): The variance of the number of patients seen per day (σ^2) is equal to 14.
Alternative hypothesis (H1): The variance of the number of patients seen per day (σ^2) is not equal to 14.

We will use a significance level of 0.01 (α = 0.01) for the test.

Step 1: Define the test statistic:

The test statistic for testing the variance is given by:

χ² = (n – 1) * s^2 / σ²

where n is the sample size, s^2 is the sample variance, and σ² is the hypothesized variance (in this case, 14).

Step 2: Calculate the test statistic:

χ² = (8 – 1) * 11.67 / 14 ≈ 6.99

Step 3: Determine the critical value:

To determine the critical value for a two-tailed test at a significance level of 0.01, we need to find the value from the chi-square distribution with 7 degrees of freedom. From statistical tables or software, we find the critical value to be approximately 20.28.

Step 4: Make a decision:

Since the test statistic (6.99) does not exceed the critical value (20.28), we fail to reject the null hypothesis.

Conclusion

Based on our analysis, we estimate with 95% confidence that the variance of the number of patients seen per day at Philip Sherman, DDS, dental clinic lies between 5.11 and 38.03. Additionally, our hypothesis test results suggest that there is not enough evidence to conclude that the variance is different from 14 at a significance level of 0.01. Thus, we do not reject the null hypothesis, indicating that there is no significant difference between the observed variance and the hypothesized variance of 14.