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. Give the lower limit.

Question

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. Give the lower limit.

Answer 1

To construct a confidence interval, we can use the formula:

Confidence Interval = sample mean ± (critical value) x (standard deviation/square root of sample size)

Given:
Sample mean (x̄) = 2.6 milligrams
Standard deviation (σ) = 0.9 milligrams
Sample size (n) = 8

First, we need to find the critical value for a 99% confidence level.

Since the sample size is small (n < 30), we use a t-distribution instead of a z-distribution.

Degrees of freedom (df) = n - 1
df = 8 - 1 = 7

Using a t-table or calculator, the critical value for a 99% confidence level with df = 7 is approximately 3.499.

Now, we can calculate the confidence interval:

Confidence Interval = 2.6 ± (3.499) x (0.9/√8)

To calculate the standard error (standard deviation/square root of sample size):
Standard Error = 0.9/√8 ≈ 0.318

Confidence Interval = 2.6 ± (3.499) x (0.318)

Calculating the values:
Lower limit = 2.6 - (3.499) x (0.318) ≈ 2.6 - 1.117 ≈ 1.483

Therefore, the lower limit of the 99% confidence interval is approximately 1.483 milligrams.

Answer 2

To construct a confidence interval, we can use the formula:

Confidence interval = sample mean ± (Critical value * Standard error)

First, let's calculate the standard error:

Standard error = standard deviation / √(sample size)

Plugging in the given values:

Standard error = 0.9 / √(8) ≈ 0.318

Next, we need to determine the critical value for a 99% confidence level. Since the sample size is small (n < 30), we can use a t-distribution instead of a z-score. The critical value can be found using a t-table or a statistical software. For a 99% confidence level with 7 degrees of freedom (n - 1), the critical value is approximately 3.499.

Now, we can compute the confidence interval:

Confidence interval = 2.6 ± (3.499 * 0.318)

Lower limit = 2.6 - (3.499 * 0.318) ≈ 1.533

Therefore, the lower limit of the 99% confidence interval for the true average nicotine content of this brand of cigarettes is approximately 1.533 milligrams.