SOLUTION: Suppose you want to test the claim that μ ≠ 3.5. Given a sample size of n = 40 and a level of α=0.05 when should you reject H 0 ? A)Reject H0 if the standardized test statisti

Algebra ->  Probability-and-statistics -> SOLUTION: Suppose you want to test the claim that μ ≠ 3.5. Given a sample size of n = 40 and a level of α=0.05 when should you reject H 0 ? A)Reject H0 if the standardized test statisti      Log On


   



Question 1158759: Suppose you want to test the claim that μ ≠ 3.5. Given a sample size of n = 40 and a level of α=0.05 when should you reject H 0 ?
A)Reject H0 if the standardized test statistic is greater than 1.645 or less than -1.645
B)Reject H0 if the standardized test statistic is greater than 2.575 or less than -2.575
C)Reject H0 if the standardized test statistic is greater than 2.33 or less than -2.33.
D)Reject H0 if the standardized test statistic is greater than 1.96 or less than -1.96.
This should be a two-tailed test, with the critical values at the α = 0.05 level of significance and n - 1 =39 degrees of freedom. The T Value calculator showed the T-Value (two-tailed)to be +/- 2.022691. I don't think I'm doing this right...

Found 2 solutions by Boreal, jim_thompson5910:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
This isn't clear to me. It is a two way test, as you have stated, but the answer choices imply that the population sd is known already, which isn't shown here. Given those choices, all of which are well-known z-values (alpha of 0.10, 0.l01, 0.01, and 0.05 respectively), the best choice would be D.
Given the information shown, your answer is best, but presumably there is a known sd in this problem for those choices.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: D) Reject H0 if the standardized test statistic is greater than 1.96 or less than -1.96.

We don't need to know the standard deviation as all we're worrying about are the boundaries to the rejection regions and the acceptance region. In other words, we aren't calculating the test statistic here.

We want to know the value of k such that P(-k < Z < k) = 0.95
The 0.95 is from 1-alpha = 1-0.05 = 0.95 which is the confidence level

Using a calculator or a Z table will show k = 1.96 approximately

Because n = 40 is larger than 30, this means we can use a standard normal Z distribution to approximate the T distribution. So we won't be using the T distribution.

A way to rephrase choice D is to say "Reject H0 if the standardized test statistic is not between -1.96 and 1.96"