How large a sample would be needed to form a 90% confidence interval for the mean nicotine content of a brand of cigarettes if the nicotine content has a normal distribution with sigma equal to 8.5 mg and the width of the interval (the length of the confidence interval) equal to 6 mg to satisfy testing requirements?
To determine the sample size needed for a 90% confidence interval, we can use the formula:
n = (Z * σ / E)^2
where:
n = sample size
Z = Z-score corresponding to the desired confidence level (in this case, 90% confidence corresponds to a Z-score of 1.645)
σ = standard deviation of the population (given as 8.5 mg)
E = maximum error tolerance (half of the width of the confidence interval)
In this case, the width of the interval is given as 6 mg. Since the confidence interval is symmetric around the mean, the maximum error tolerance would be half of that, which is 3 mg.
Plugging in the values into the formula:
n = (1.645 * 8.5 / 3)^2
n ≈ 4.58^2
n ≈ 21
Therefore, a sample size of at least 21 cigarettes is needed to form the 90% confidence interval for the mean nicotine content of the brand.