Respuesta :
Answer:
[tex]\mu \leq 700 + t_{(0.025,15)}\frac{50}{4}[/tex], 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 [tex]\alpha[/tex] level of [tex]\frac{1 - 0.95}{2} = 0.025[/tex], thus, we have [tex]t_{(0.025,15)}[/tex]
The margin of error is of:
[tex]M = t_{(0.025,15)}\frac{s}{\sqrt{n}}[/tex]
In which s is the standard deviation of the sample and n is the size of the sample. For this question, we have [tex]s = 50,n = 16[/tex]. So
[tex]M = t_{(0.025,15)}\frac{s}{\sqrt{n}}[/tex]
[tex]M = t_{(0.025,15)}\frac{50}{\sqrt{16}}[/tex]
[tex]M = t_{(0.025,15)}\frac{50}{4}[/tex]
Upper bound:
The upper bound is less than the sample mean of 700 added to the margin of error of M. Thus
[tex]\mu \leq 700 + t_{(0.025,15)}\frac{50}{4}[/tex]
And the correct answer is given by option c.
