Question

sample of size n = 8 from a Normal(1, 0) population results in a sample standard deviation of s = 5.4. A 95% lower bound for the true population standard deviation is A: 0 > 1.016. B: 0 > 2.687. C: 0 > 0.384 D: 0 > 1.783. E: o > 3.809

Answer 1

Explanation:

To calculate a lower bound for the population standard deviation with a 95% confidence level, we can use the following formula:

lower bound = (n - 1) * s^2 / chi-squared(alpha/2, n-1)

where n is the sample size, s is the sample standard deviation, chi-squared(alpha/2, n-1) is the chi-squared value at the alpha/2 percentile with n-1 degrees of freedom.

In this case, n = 8 and s = 5.4. For a 95% confidence level, alpha/2 = 0.025, so we need to find the chi-squared(0.025, 7) value.

Using a chi-squared table or calculator, we find that chi-squared(0.025, 7) = 14.0671.

Plugging in the values, we get:

lower bound = (8 - 1) * 5.4^2 / 14.0671

= 19.4175

Taking the square root of this value gives us the lower bound for the population standard deviation:

sqrt(19.4175) = 4.4084

Therefore, the correct answer is E: o > 3.809, as the lower bound for the population standard deviation is greater than 3.809.