. A random sample of 9 cigarettes has an average nicotine content of 2.5 milligram with a standard deviation of 0.8 milligram. Construct a 95% confidence interval for the true average nicotine content of this brand of cigarettes, assuming the distribution of nicotine contents to be approximately normal.

1 answer

To construct a 95% confidence interval for the true average nicotine content, we can use the formula for the confidence interval based on the sample mean, sample standard deviation, and the t-distribution since the sample size is small (n < 30).

Step 1: Find the sample mean (\( \bar{x} \)) and sample standard deviation (\( s \)):

  • Sample mean (\( \bar{x} \)) = 2.5 mg
  • Sample standard deviation (\( s \)) = 0.8 mg
  • Sample size (\( n \)) = 9

Step 2: Determine the t-value for a 95% confidence level:

Since the degrees of freedom is \( n - 1 \) (where \( n \) is the sample size):

\[ \text{Degrees of freedom} = n - 1 = 9 - 1 = 8 \]

Using a t-table or calculator, we find the t-value for \( df = 8 \) at a 95% confidence level is approximately \( t_{0.025, 8} \approx 2.306 \).

Step 3: Calculate the standard error (SE):

The standard error (SE) is calculated using the sample standard deviation and the sample size:

\[ SE = \frac{s}{\sqrt{n}} = \frac{0.8}{\sqrt{9}} = \frac{0.8}{3} \approx 0.267 \]

Step 4: Calculate the margin of error (ME):

The margin of error (ME) is calculated by multiplying the critical t-value by the standard error:

\[ ME = t \times SE = 2.306 \times 0.267 \approx 0.616 \]

Step 5: Construct the confidence interval:

To find the confidence interval, we take the sample mean and add/subtract the margin of error:

\[ \text{Lower limit} = \bar{x} - ME = 2.5 - 0.616 \approx 1.884 \]

\[ \text{Upper limit} = \bar{x} + ME = 2.5 + 0.616 \approx 3.116 \]

Conclusion:

The 95% confidence interval for the true average nicotine content of this brand of cigarettes is approximately:

\[ \text{Confidence Interval} = (1.884 \text{ mg}, 3.116 \text{ mg}) \]