Question

Which critical value is appropriate for a​ 99% confidence level where nequals​17, sigma is​ unknown, and the population appears to be normally​ distributed? A. z Subscript alpha divided by 2equals2.567 B. t Subscript alpha divided by 2equals2.921 C. z Subscript alpha divided by 2equals2.583 D. t Subscript alpha divided by 2equals2.898

Answer 1

Answer:

t Subscript alpha divided by 2 = 2.921

Step-by-step explanation:

With sample sizes lower than 30, we use the t-test.

Otherwise, we use the z-test.

Here, n = 17

So we use the t-test.

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 = 17 - 1 = 16

Now, we have to find a value of T, which is found looking at the t table, with 16 degrees of freedom(y-axis) and a confidence level of 0.99($t_{1 - (\frac{1-0.99}{2}}) = 0.995$). So we have T = 2.921

So the correct answer is:

t Subscript alpha divided by 2 = 2.921

Answer 2

Answer:

Critical value that is appropriate for a​ 99% confidence level is 2.921.

Step-by-step explanation:

We are given that n = 17, sigma is​ unknown, and the population appears to be normally​ distributed.

We have to determine that which critical value is appropriate for a​ 99% confidence level.

So, firstly we will decide that which table should be used for looking for the critical value;

As we know that;

• z-score table is used when we have information about the population standard deviation (sigma) means we know the value of sigma.

• t-table is used when we don't have information about the population standard deviation (sigma) means we don't know the value of sigma.

So, here in this question the sigma is unknown, which means we will consider t-table here.

Now, we will come to what degree of freedom to choose. The t-table has a degree of freedom of n - 1, i.e. 17 - 1 = 16.

Also, in t-table there is a variable P which is  $\frac{\alpha}{2}$ where $\alpha$ is the significance level.

In this question we have significance level of 1%, so P = $\frac{1}{2}$ = 0.5%.

Now, in table we will look for the value where P = 0.5% and degree of freedom = 16 which will give us the value of 2.921.

Therefore, critical value that is appropriate for a​ 99% confidence level is 2.921.