A sample size of 5 provides a sample mean of 9.6. If the population variance is known to be 5 and the population distribution is assumed to be normal, the lower limit for a 92% confidence interval is 7.85.

Question

Quizplus · Accepted Answer

To find the 92% confidence interval, we need to use the formula:
CI = X̄ ± Zα/2 * σ/√n
where:
X̄ = sample mean = 9.6
Zα/2 = the z-score associated with a 92% confidence level = 1.75 (from z-tables)
σ = population standard deviation = √5 = 2.236
n = sample size = 5

Plugging in the values, we get:
CI = 9.6 ± 1.75 * 2.236/√5
CI = (7.856, 11.344)

The lower limit of this interval is 7.856, which is greater than the given lower limit of 7.85. Therefore, the statement is false.

A Sample Size of 5 Provides a Sample Mean of 9.6