A group of eight individuals with high cholesterol levels were given a new drug that was designed to lower cholesterol levels. Cholesterol levels, in milligrams per deciliter, were measured before and after treatment for each individual, with the following results:

Question

Individual 1- before: 258 , after: 195
individual 2- 260 / 180
3- 280 / 183
4- 265 / 187
5- 291 / 170
6- 281 / 203
7- 281 / 179
8- 285 / 185
Construct a 90% confidence interval for the mean reduction in cholesterol level. Round to 1 decimal point.

Answer 1

To construct a 90% confidence interval for the mean reduction in cholesterol level, we will follow these steps:

1. Calculate the reductions for each individual.
2. Calculate the mean and standard deviation of the reductions.
3. Determine the confidence interval.

Step 1: Calculate the Reductions

For each individual, the reduction is calculated as: Reduction = Before - After

The reductions for the individuals are:

• Individual 1: \(258 - 195 = 63\)
• Individual 2: \(260 - 180 = 80\)
• Individual 3: \(280 - 183 = 97\)
• Individual 4: \(265 - 187 = 78\)
• Individual 5: \(291 - 170 = 121\)
• Individual 6: \(281 - 203 = 78\)
• Individual 7: \(281 - 179 = 102\)
• Individual 8: \(285 - 185 = 100\)

So the reductions are:

• 63
• 80
• 97
• 78
• 121
• 78
• 102
• 100

Step 2: Calculate the Mean and Standard Deviation of the Reductions

The mean reduction (\( \bar{x} \)) can be calculated as: \[ \bar{x} = \frac{\sum \text{reductions}}{n} \] where \(n\) is the number of individuals.

Calculating the sum of reductions: \[ 63 + 80 + 97 + 78 + 121 + 78 + 102 + 100 = 719 \]

Number of individuals \(n = 8\).

Now calculate the mean: \[ \bar{x} = \frac{719}{8} = 89.875 \]

Next, we calculate the standard deviation (\(s\)): \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \]

Calculating each squared deviation:

• \( (63 - 89.875)^2 = (-26.875)^2 = 722.265625 \)
• \( (80 - 89.875)^2 = (-9.875)^2 = 97.515625 \)
• \( (97 - 89.875)^2 = (7.125)^2 = 50.640625 \)
• \( (78 - 89.875)^2 = (-11.875)^2 = 141.640625 \)
• \( (121 - 89.875)^2 = (31.125)^2 = 967.265625 \)
• \( (78 - 89.875)^2 = (-11.875)^2 = 141.640625 \)
• \( (102 - 89.875)^2 = (12.125)^2 = 147.265625 \)
• \( (100 - 89.875)^2 = (10.125)^2 = 102.515625 \)

Sum of squared deviations: \[ 722.265625 + 97.515625 + 50.640625 + 141.640625 + 967.265625 + 141.640625 + 147.265625 + 102.515625 = 2272.375 \]

Now calculate the standard deviation: \[ s = \sqrt{\frac{2272.375}{8-1}} = \sqrt{\frac{2272.375}{7}} = \sqrt{324.61} \approx 18.0 \]

Step 3: Determine the Confidence Interval

For a 90% confidence interval with \(n-1\) degrees of freedom (7 degrees of freedom), we can find the critical t-value. Let's use the t-table:

The t-value for 90% confidence and 7 degrees of freedom is approximately \( t_{0.05} \approx 1.895\).

Now, we calculate the margin of error (ME): \[ ME = t \times \frac{s}{\sqrt{n}} = 1.895 \times \frac{18.0}{\sqrt{8}} = 1.895 \times \frac{18.0}{2.828} \approx 1.895 \times 6.364 = 12.07 \]

Finally, we calculate the confidence interval: \[ \text{Lower limit} = \bar{x} - ME \approx 89.875 - 12.07 \approx 77.8 \] \[ \text{Upper limit} = \bar{x} + ME \approx 89.875 + 12.07 \approx 101.9 \]

Conclusion

The 90% confidence interval for the mean reduction in cholesterol level is approximately:

\[ \text{(77.8, 101.9)} \]

Answer 2

(b) A physician claims that the mean reduction in cholesterol level is more than 50 milligrams per deciliter. Does the confidence interval contradict this claim?
[yes or no?]. the confidence interval [does or doesnt] contradict the claim

Answer 3

To answer this question, let's revisit the 90% confidence interval for the mean reduction in cholesterol level that we calculated previously, which is approximately:

\[ (77.8, 101.9) \]

The physician claims that the mean reduction in cholesterol level is more than 50 milligrams per deciliter.

Now, let's evaluate the confidence interval:

• The entire confidence interval (77.8, 101.9) is above 50.

Conclusion:

• Since the entire confidence interval is above 50, it does not contradict the physician's claim.

Final Answer:

Yes, the confidence interval doesn't contradict the claim.

Answer 4

your answer is confusing, does the confidence interval contradict the claim

Answer 5

I apologize for any confusion. Let's clarify:

The confidence interval we calculated is approximately \( (77.8, 101.9) \).

The physician's claim is that the mean reduction in cholesterol level is more than 50 mg/dL.

Since the entire confidence interval lies above 50 (both \( 77.8 \) and \( 101.9 \) are greater than \( 50 \)), it supports the physician's claim rather than contradicting it.

Clear Answer:

No, the confidence interval does not contradict the claim.