Question

A news report states that the 95% confidence interval for the mean number of daily calories consumed by participants in a medical study is (1960, 2170). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 15 observations. Calculate the sample mean, the margin of error, and the sample standard deviation based on the stated confidence interval and the given sample size. Use the t distribution in any calculations and round non-integer results to 4 decimal places.1. What is the sample mean? 2. What is the margin of error? 3. What is the sample standard deviation?

Answer 1

Answer: 1) sample mean = 2065

2) margin of error = 105

3) sample standard deviation = 48.95

Explanation:

Confidence interval is written in the form,

(Point estimate - margin of error, Point estimate + margin of error)

The sample mean, x is the point estimate for the population mean. Let m represent the margin of error. Since the confidence interval is given as (1960 , 2170), it means that

x - m = 1960

x + m = 2170

Adding both equations, it becomes

2x = 4130

x = 4130/2

x = 2065

Substituting x = 2065 into x + m = 2170, it becomes

2065 + m = 2170

m = 2170 - 2065

m = 105

Using the t distribution,

Degree of freedom, df = 15 - 1 = 14

α = 1 - 0.95 = 0.05

z α/2 = 0.05/2 = 0.025

The area to the right of z 0.025 is 0.05 and the area to the left of z0.025 is 1 - 0.025 = 0.975. Using the t distribution table,

z = 2.145

Confidence interval = mean ± z × σ/√n

Where

σ = population standard Deviation

σ/√n = sample standard deviation

Confidence interval = sample mean(point estimate) ± z × σ/√n

Considering the lower boundary of the confidence interval,

1960 = 2065 - 2.145 × σ/√n

1960 - 2065 = - 2.145 × σ/√n

- 105 = - 2.145 × σ/√n

σ/√n = - 105/- 2.145

σ/√n = 48.95