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 (1850, 2090). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 22 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.

Answer 1

Answer:

The sample mean is of 1970 daily calories.

The margin of error is of 120 daily calories.

The sample standard deviation is of 12.3 daily calories.

Step-by-step explanation:

Mean number of daily calories consumed by participants in a medical study is (1850, 2090).

The sample mean is the mean of the bounds of the confidence interval. So

$M = \frac{1850 + 2090}{2} = 1970$

The sample mean is of 1970 daily calories.

Margin of error:

The margin of error is the difference between the bounds and the sample mean. So

2090 - 1970 = 1970 - 1850 = 120

The margin of error is of 120 daily calories.

Sample standard deviation:

Now I have to expand a bit into the t-distribution. So

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 22 - 1 = 21

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 21 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 2.08

The margin of error is:

$M = T\frac{s}{\sqrt{n}}$

In which s is the standard deviation of the sample.

Since $M = 120, T = 2.08, n = 22$

$120 = 2.08\frac{s}{\sqrt{22}}$

$s = \frac{120}{2.08\sqrt{22}} = 12.3$

The sample standard deviation is of 12.3 calories.