Eleven employees were put under the care of the company nurse because of high cholesterol readings. Construct a 98% confidence interval to estimate the population mean difference o... Eleven employees were put under the care of the company nurse because of high cholesterol readings. Construct a 98% confidence interval to estimate the population mean difference of cholesterol readings for people involved in this program.
Understand the Problem
The question is asking to construct a 98% confidence interval for the mean difference in cholesterol readings of 11 employees before and after a new diet program. It highlights the need to analyze data normally distributed within the population.
Answer
The confidence interval for the mean difference in cholesterol readings is \( \left( \bar{D} - E, \bar{D} + E \right) \).
Answer for screen readers
The 98% confidence interval for the mean difference in cholesterol readings is ( \left( \bar{D} - E, \bar{D} + E \right) ).
Steps to Solve
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Collect Data and Calculate Differences
Gather the cholesterol readings for each employee both before and after the diet. Calculate the differences between the after and before readings for each employee.
Let ( D_i ) represent the difference for each employee ( i ): [ D_i = \text{Post-Diet Reading} - \text{Pre-Diet Reading} ]
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Find the Mean and Standard Deviation of Differences
Calculate the mean (( \bar{D} )) and standard deviation (( s_D )) of the differences. The formulas are: [ \bar{D} = \frac{\sum_{i=1}^{n} D_i}{n} ] [ s_D = \sqrt{\frac{\sum_{i=1}^{n} (D_i - \bar{D})^2}{n-1}} ] where ( n ) is the number of employees.
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Determine the Critical Value
For a 98% confidence interval and 10 degrees of freedom (since ( n - 1 = 11 - 1 = 10 )), use the t-distribution table to find the critical value ( t^* ).
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Calculate the Margin of Error
The margin of error (( E )) can be calculated using the formula: [ E = t^* \cdot \frac{s_D}{\sqrt{n}} ]
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Construct the Confidence Interval
The 98% confidence interval for the mean difference is given by: [ \left( \bar{D} - E, \bar{D} + E \right) ]
The 98% confidence interval for the mean difference in cholesterol readings is ( \left( \bar{D} - E, \bar{D} + E \right) ).
More Information
The confidence interval provides a range in which we can be 98% certain that the true mean difference in cholesterol readings lies. This is particularly important for making informed decisions based on the diet program's effectiveness.
Tips
- Not calculating the differences correctly, which can lead to incorrect mean and standard deviation.
- Using a z-score instead of a t-score for smaller sample sizes (n < 30).
- Failing to check the assumptions of normality when applying the t-distribution.
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