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A healthcare provider monitors the number of CAT scans performed each month in each of its clinics. The most recent year of data for a particular clinic follows (the reported variable is the number of CAT scans each month expressed as the number of CAT scans per thousand members of the health plan): \(2.31,2.09,2.36,1.95,1.98,2.25,2.16,2.07,1.88,1.94,1.97,\) \(2.02 .\) (a) Find a \(95 \%\) two-sided CI on the mean number of CAT scans performed each month at this clinic. (b) Historically, the mean number of scans performed by all clinics in the system has been \(1.95 .\) If there any evidence that this particular clinic performs more CAT scans on average than the overall system average?

Short Answer

Expert verified
CI is (2.03, 2.23). There is evidence that this clinic performs more scans than the system average.

Step by step solution

01

Calculate the Mean

First, we sum up all the monthly CAT scans and divide by the number of months to find the mean: \[ \text{Mean} = \frac{2.31 + 2.09 + 2.36 + 1.95 + 1.98 + 2.25 + 2.16 + 2.07 + 1.88 + 1.94 + 1.97 + 2.02}{12} \approx 2.13 \]
02

Calculate the Standard Deviation

To find the standard deviation, we first subtract the mean from each month's value, square the result, then find the average of those squares, and finally take the square root. Calculating each deviation and its square: - \((2.31 - 2.13)^2 \approx 0.0324\) - \((2.09 - 2.13)^2 \approx 0.0016\) - ... The sum of these squared deviations is approximately 0.2772. Standard deviation (s) is calculated as: \[ s = \sqrt{\frac{0.2772}{12 - 1}} \approx 0.158 \]
03

Find the Critical t-value

For a 95% confidence interval with 11 degrees of freedom ( n = 12 - 1 df = 11 df = 11 df = 11, n-1, consult a t-table), the critical t-value is approximately 2.201.
04

Calculate the Confidence Interval

Now, we use the formula for the confidence interval:\[ \text{CI} = \bar{x} \pm t \left(\frac{s}{\sqrt{n}}\right) \]Where \(\bar{x}\)mean= 2.13, \(t\)critical value= 2.201, \(s\)standard deviation= 0.158, and n = 12.\[ \text{CI} = 2.13 \pm 2.201 \left(\frac{0.158}{\sqrt{12}}\right) \approx 2.13 \pm 0.100 \]The confidence interval is approximately (2.03, 2.23).
05

Hypothesis Testing

We need to test if the mean number of scans in this clinic is significantly greater than 1.95. Use: - Null hypothesis \(H_0: \mu = 1.95\)- Alternative hypothesis \(H_a: \mu > 1.95\)Calculate the t-statistic:\[ t = \frac{2.13 - 1.95}{0.158 / \sqrt{12}} \approx 3.83 \]
06

Compare t-statistic with t-value

The calculated t-statistic is 3.83. For a one-tailed test with df = 11 = 11 = 11,= 11, t=11,= 11 degrees of freedom and significance level \(\alpha = 0.025\), the critical t-value is approximately 1.796.Since 3.83 > 1.796, we reject the null hypothesis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

t-distribution
The t-distribution is a theoretical probability distribution that is symmetrical and bell-shaped, much like the normal distribution, but it has heavier tails. This means it is more spread out. The t-distribution is particularly important when dealing with small sample sizes, typically n < 30. It is used when the population standard deviation is unknown, and instead, the sample standard deviation is used.

In the context of confidence intervals and hypothesis testing, the t-distribution allows us to calculate critical values or t-values which are necessary to determine the range in which our population parameter likely falls. Each t-distribution is defined by its degrees of freedom, which are related to the sample size. As the sample size increases, the t-distribution approaches a normal distribution.
  • Critical t-value: The point on the t-distribution that marks the boundary of the confidence interval for a given confidence level and degrees of freedom.
  • Degrees of freedom: Calculated as n - 1 for a single sample, where n is the sample size. They adjust the t-distribution to better fit the variability expected in small samples.
Therefore, we use a t-distribution in the exercise to find a 95% confidence interval for the mean number of CAT scans, as the sample is small and the standard deviation of the population is not known.
hypothesis testing
Hypothesis testing involves making an assumption about a population parameter and testing whether this assumption is true. It's a method we use to draw conclusions about the population based on sample data. Hypothesis testing usually involves two hypotheses: the null hypothesis ( H_0 ) and the alternative hypothesis ( H_a ). The null hypothesis is generally an assumption of 'no effect' or 'no difference,' while the alternative hypothesis is what we aim to support.

In this exercise, the null hypothesis is that the mean number of CAT scans is equal to the system average, which is 1.95. The alternative hypothesis proposes that the clinic performs more CAT scans than the system average.
  • Null Hypothesis ( H_0 ): Suggests no difference or effect. In our case, it is that the mean number of scans is 1.95.
  • Alternative Hypothesis ( H_a ): Suggests the presence of a difference or effect. Here, it is that the mean number of scans is greater than 1.95.
  • t-statistic: A calculated value using sample data to test the hypothesis. If this value exceeds the critical t-value from the t-distribution, the null hypothesis is rejected.
This approach provides a structured framework to confirm if the observed sample data provides significant evidence against the null hypothesis.
standard deviation
Standard deviation is a measure of the amount of variation or dispersion within a set of values. It tells us how much the individual data points deviate from the mean on average. A smaller standard deviation indicates that data points are closer to the mean, whereas a larger one suggests more spread-out data.

The standard deviation is crucial when calculating confidence intervals because it influences the width of the interval. In the exercise provided, the sample standard deviation is essential for calculating the margin of error and subsequently the confidence interval.
  • Formula to calculate sample standard deviation (s): \(s = \sqrt{\frac{1}{n-1}\sum (x_i - \bar{x})^2}\), where \(x_i\) are the individual data points and \(\bar{x}\) is the mean.
  • Understand the spread: A low s indicates less variation in monthly CAT scans, suggesting consistent monthly totals.
The calculated standard deviation of approximately 0.158 for the clinic tells us about the variability in the number of CAT scans across months and helps us interpret the reliability of the mean.

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Most popular questions from this chapter

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