Question

A simple random sample of 10 households, the number of TVs that each household had is as follows:

1, 1, 2, 2, 4, 4, 1, 3, 5, 5
Assume that it is reasonable to believe that the population is approximately normal, and the population standard deviation is 0.40. What is the lower bound of the 95% confidence interval for the mean number of TVs?

Answer 1

The lower bound of the 95% confidence interval for the mean number of TVs is approximately 2.553.

Step-by-step explanation:

To calculate the lower bound of the 95% confidence interval for the mean number of TVs, we can use the formula: Lower bound = sample mean - (critical value * standard deviation / sqrt(sample size)).

Given that we have a simple random sample of 10 households with a sample mean of 2.8 and a population standard deviation of 0.40, we need to find the critical value for a 95% confidence level. By referring to the Z-table, we can determine that the critical value is approximately 1.96. Plugging in these values into the formula, we get:

Lower bound = 2.8 - (1.96 * 0.40 / sqrt(10))

Lower bound = 2.8 - (1.96 * 0.126)

Lower bound = 2.8 - 0.247

Lower bound = 2.553

Therefore, the lower bound of the 95% confidence interval for the mean number of TVs is approximately 2.553.