2. Suppose the 90% confidence interval for the mean SAT scores of applicants at a business college is given by [1690, 1810]. This confidence interval uses the sample mean and sample standard deviation based on 25 observations. What are the sample mean and sample standard deviation used for this interval

Respuesta :

Answer:

The sample mean used for this interval is 1750.

The sample standard deviation used for this interval was of 175.34

Step-by-step explanation:

Confidence interval concepts:

A confidence interval has two bounds, a lower bound and an upper bound.

A confidence interval is symmetric, which means that the point estimate used is the mid point between these two bounds, that is, the mean of the two bounds.

The margin of error is the subtraction of these two bounds divided by two.

In this question:

Lower bound: 1690

Upper bound: 1810

Sample mean

[tex]\frac{1690 + 1810}{2} = 1750[/tex]

The sample mean used for this interval is 1750.

Sample standard deviation:

The first step is finding the margin of error:

[tex]M = \frac{1810 - 1690}{2} = 60[/tex]

Now we have to develop the problem a bit.

We want the sample standard deviation, so we use the T-distribution.

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 = 25 - 1 = 24

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 1.711

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}}[/tex]

We have that: [tex]M = 60, T = 1.711, n = 25[/tex]

We have to find s

[tex]M = T\frac{s}{\sqrt{n}}[/tex]

[tex]60 = 1.711\frac{s}{\sqrt{25}}[/tex]

[tex]1.711s = 60*5[/tex]

[tex]s = \frac{60*5}{1.711}[/tex]

[tex]s = 175.34[/tex]

The sample standard deviation used for this interval was of 175.34