Question

According to Lear Center Local News Archive, the average amount of time that a half-hour local TV

news broadcast devotes to U.S. foreign policy, including the war in Iraq, is 38 seconds with a standard
deviation of 8 seconds. (Time, February 28, 2005). Suppose a random sample of 35 such half-hour
news broadcasts shows that an average of 36 seconds are devoted to U.S. foreign policy. Find a 95%
confidence interval for the mean time that all half-hour local TV news broadcasts devote to U.S.
foreign policy to corroborate or refute the claim.
Round to one decimal place.
Does the confidence interval suppose the Lear Center's claim?

Answer 1

Explanation:

We are given:

Sample size (n) = 35

Sample mean ($\bar{x}$) = 36 seconds

Population standard deviation ($\sigma$) = 8 seconds

Confidence level = 95%

We can find the margin of error using the formula:

Margin of Error = z*(σ/√n), where z is the z-score for the given confidence level, σ is the population standard deviation, and n is the sample size.

For a 95% confidence level, the z-score is 1.96 (from the z-table or calculator).

Putting the values in the formula, we get:

Margin of Error = 1.96*(8/√35) ≈ 2.68

The confidence interval is given by:

Lower Limit = $\bar{x}$ - Margin of Error

Upper Limit = $\bar{x}$ + Margin of Error

Substituting the values, we get:

Lower Limit = 36 - 2.68 ≈ 33.32

Upper Limit = 36 + 2.68 ≈ 38.68

Therefore, the 95% confidence interval for the mean time that all half-hour local TV news broadcasts devote to U.S. foreign policy is (33.32, 38.68).

Since the Lear Center's claim of the average time being 38 seconds falls within this interval, we can say that the sample data supports the Lear Center's claim at a 95% confidence level.