/ 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 90% 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 90% confidence interval for the true average nicotine content of the cigarettes, we can use the formula for the confidence interval (CI) when the sample size is small (n < 30) and the population standard deviation is unknown. In this case, we will use the t-distribution.

  1. Identify the sample statistics:

    • Sample mean (\(\bar{x}\)) = 2.6 mg
    • Sample standard deviation (s) = 0.9 mg
    • Sample size (n) = 8
  2. Determine the critical value: Since the confidence level is 90%, we can find the critical value \( t \) for a 90% confidence level and 7 degrees of freedom (n - 1 = 8 - 1 = 7).

Using a t-table or calculator, we find:

  • The critical value \( t \) for 90% confidence level with 7 degrees of freedom is approximately 1.895.
  1. Calculate the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} = \frac{0.9}{\sqrt{8}} \approx \frac{0.9}{2.828} \approx 0.3182 \]

  2. Calculate the margin of error (ME): \[ ME = t \times SE = 1.895 \times 0.3182 \approx 0.603 \]

  3. Construct the confidence interval: \[ CI = \bar{x} \pm ME = 2.6 \pm 0.603 \] This gives us:

    • Lower limit: \( 2.6 - 0.603 \approx 1.997 \)
    • Upper limit: \( 2.6 + 0.603 \approx 3.203 \)

Thus, the 90% confidence interval for the true average nicotine content of this brand of cigarettes is approximately (1.997 mg, 3.203 mg).