Question

A nutritionist wants to estimate the average caloric content of the burritos at a popular restaurant. They obtain a random sample of 16 burritos and measure their caloric content. The sample mean and sample standard deviation of measured caloric contents of the 16 burritos are 700 calories and 50 calories, respectively. Suppose the caloric content of burritos is normally distributed. Based on the sample, which of the following is a 95% upper confidence bound for the mean caloric content of these burritos?
A. μ≤700+ t 0.05, 15x50/4
B. μ≤700+ t 0.025, 16x50/4
C. μ≤700+ t 0.025, 15x50/4
D. μ≤700+ z 0.05 x50/4

Answer 1

Answer:

$\mu \leq 700 + t_{(0.025,15)}\frac{50}{4}$, option c

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 = 16 - 1 = 15

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 15 degrees of freedom(y-axis) and an $\alpha$ level of $\frac{1 - 0.95}{2} = 0.025$, thus, we have $t_{(0.025,15)}$

The margin of error is of:

$M = t_{(0.025,15)}\frac{s}{\sqrt{n}}$

In which s is the standard deviation of the sample and n is the size of the sample. For this question, we have $s = 50,n = 16$. So

$M = t_{(0.025,15)}\frac{s}{\sqrt{n}}$

$M = t_{(0.025,15)}\frac{50}{\sqrt{16}}$

$M = t_{(0.025,15)}\frac{50}{4}$

Upper bound:

The upper bound is less than the sample mean of 700 added to the margin of error of M. Thus

$\mu \leq 700 + t_{(0.025,15)}\frac{50}{4}$

And the correct answer is given by option c.