Question

A nurse is investigating the birth weights of​ non-premature babies. She selects a random sample of 41 such​ babies, and found a mean weight of 3506.4 grams and a sample standard deviation of
494.2 grams. Find a 95​% confidence interval for the mean birth weight of all ​non-premature babies. Show your calculation for the confidence interval as well as the final confidence interval rounded to one decimal place. Then write a sentence that gives the interpretation of your confidence​ interval, including the correct units.

Answer 1

Answer:

The 95​% confidence interval for the mean birth weight of all ​non-premature babies is between 3350.4 grams and 3662.4 grams. This means that we are 95% sure that true mean birth weight of all non-premature babies is in this interval.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

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 = 41 - 1 = 40

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 40 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 2.0211

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.0211\frac{494.2}{\sqrt{41}} = 156$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 3506.4 - 156 = 3350.4 grams

The upper end of the interval is the sample mean added to M. So it is 3506.4 + 156 = 3662.4 grams

The 95​% confidence interval for the mean birth weight of all ​non-premature babies is between 3350.4 grams and 3662.4 grams. This means that we are 95% sure that true mean birth weight of all non-premature babies is in this interval.