Question

Construct a 90% confidence interval for a sampling distribution having means of {4, 4,

4, 4.2, 4.2, 4.3, 4.3, 4.3, 4.4, 4.4, 4.4, 4.4, 4.5, 4.5, 4.6, 4.7, 4.7, 4.7, 4.8, 4.8, 4.8, 4.9, 4.9,
4.9, 4.9, 5, 5, 5, 5, 5, 5, 5.1, 5.1, 5.1, 5.2, 5.2). The standard deviation of the population
from which these samples were drawn was known to be a = 0.357.
DOC
4.567-4.783
4.577-4.773
4.558-4.792
10-90%

Answer 1

To construct a 90% confidence interval, we can use the formula: CI = x +- z(alpha/2) * sigma/root(n), where CI represents the confidence interval, x is the sample mean, z(alpha/2) is the critical value, sigma is the population standard deviation, and n is the sample size. Plugging in the given values, we find that the 90% confidence interval is (4.224, 4.576).

Step-by-step explanation:

To construct a 90% confidence interval, we can use the formula:

CI = x ± z(α/2) * σ/√n

where CI represents the confidence interval, x is the sample mean, z(α/2) is the critical value corresponding to the desired level of confidence, σ is the population standard deviation, and n is the sample size.

Plugging in the given values, we have:

CI = 4.4 ± (1.645) * (0.357 / √34)

Simplifying this expression, we get:

CI = 4.4 ± 0.176

Therefore, the 90% confidence interval for the given sampling distribution is (4.224, 4.576).