A random sample of 8 cigarettes of a certain brand has an average nicotine content of 2.6 milligram and standard deviation of 0.9 milligrams. Construct a 99% confidence interval for the true average nicotine content of this brand of cigarettes, assuming the distribution of nicotine content to be normal.

1 answer

To construct a 99% confidence interval for the true average nicotine content, we can use the formula for the confidence interval for the mean when the population standard deviation is unknown. This requires us to use the t-distribution since our sample size is small (n < 30).

The formula for the confidence interval is given by:

\[ \bar{x} \pm t_{\frac{\alpha}{2}} \left(\frac{s}{\sqrt{n}}\right) \]

Where:

  • \(\bar{x}\) is the sample mean
  • \(t_{\frac{\alpha}{2}}\) is the t-value for a 99% confidence level
  • \(s\) is the sample standard deviation
  • \(n\) is the sample size

Given:

  • Sample mean (\(\bar{x}\)) = 2.6 mg
  • Sample standard deviation (\(s\)) = 0.9 mg
  • Sample size (\(n\)) = 8
  • Confidence level = 99%

Step 1: Find \(t_{\frac{\alpha}{2}}\)

  1. Calculate degrees of freedom (df): \[ df = n - 1 = 8 - 1 = 7 \]

  2. Find the t-value: For a 99% confidence level, the alpha level (\(\alpha\)) is 0.01. Therefore, \(\frac{\alpha}{2} = 0.005\). We can look up the t-value for 7 degrees of freedom.

Using a t-table or calculator, we find: \[ t_{0.005, 7} \approx 3.499 \]

Step 2: Calculate the standard error (SE)

The standard error (SE) is calculated as follows: \[ SE = \frac{s}{\sqrt{n}} = \frac{0.9}{\sqrt{8}} \approx \frac{0.9}{2.828} \approx 0.318 \]

Step 3: Calculate the margin of error (ME)

The margin of error (ME) is: \[ ME = t_{\frac{\alpha}{2}} \times SE = 3.499 \times 0.318 \approx 1.113 \]

Step 4: Construct the confidence interval

Now we can construct the confidence interval: \[ \text{Confidence Interval} = \bar{x} \pm ME = 2.6 \pm 1.113 \]

Calculating the lower and upper limits:

  • Lower limit: \(2.6 - 1.113 \approx 1.487\)
  • Upper limit: \(2.6 + 1.113 \approx 3.713\)

Final Result

The 99% confidence interval for the true average nicotine content of this brand of cigarettes is approximately:

\[ (1.487, 3.713) \]

This means we can be 99% confident that the true average nicotine content falls within this interval.