Question

in a study of sodium intake, a sample of u.s residents was found to have a mean daily sodium intake of 3,720 mg. suppose that this result was based on a sample of size 26 and that the sample standard deviation is 1,148 mg. determine the 98% confidence interval for the true mean of daily sodium intake.

Answer 1

To determine the 98% confidence interval for the true mean of daily sodium intake, use the formula CI = x ± Z * (σ / √n), and plug in the given values for x, Z, σ, and n. The resulting confidence interval is approximately (3173.17, 4266.83) mg.

Step-by-step explanation:

To determine the 98% confidence interval for the true mean of daily sodium intake, we can use the formula:

CI = x ± Z * (σ / √n)

Where:

• CI is the confidence interval
• x is the sample mean
• Z is the Z-score corresponding to the desired confidence level (in this case, 98%)
• σ is the sample standard deviation
• n is the sample size

Plugging in the values given:

x = 3720 mg

Z = 2.33 (corresponding to a 98% confidence level)

σ = 1148 mg

n = 26

Substituting these values into the formula, we get:

CI = 3720 ± 2.33 * (1148 / √26)

Simplifying the expression:

CI ≈ 3720 ± 546.83

Thus, the 98% confidence interval for the true mean of daily sodium intake is approximately (3173.17, 4266.83) mg.