Mathematics, 05.03.2020 11:44 rubianny03

A simple random sample of 26 precipitation amounts has a standard deviation of 0.18. Find the test statistic, using a 0.01 significance level to test a claim that the standard deviation of all precipitation amounts is equal to 0.25

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Answer from: ash9951

$t=(26-1) [\frac{0.18}{0.25}]^2 =12.96$

What is the critical value for the test statistic at an α = 0.01 significance level?

Since is a two tailed test the critical zone have two zones. On this case we need a quantile on the chi square distribution with 25 degrees of freedom that accumulates 0.005 of the area on the left tail and 0.995 on the right tail.

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.005,25)". And our critical value would be $\Chi^2 =10.520$

And the right critical value would be : $\Chi^2 =46.927$

And the rejection zone would be: $\chi^2 < 10.52 \cup \chi^2 46.927$

Since our calculated value is NOT in the rejection zone we FAIL to reject the null hypothesis.

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

n = 26 sample size

s= 0.18

$\sigma_o =0.25$ the value that we want to test

p_v represent the p value for the test

t represent the statistic

$\alpha=0.01$ significance level

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is equal to 0.25, so the system of hypothesis are:

H0: $\sigma =0.25$

H1: $\sigma \neq 0.25$

In order to check the hypothesis we need to calculate the statistic given by the following formula:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

$t=(26-1) [\frac{0.18}{0.25}]^2 =12.96$

What is the critical value for the test statistic at an α = 0.01 significance level?

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.005,25)". And our critical value would be $\Chi^2 =10.520$

And the right critical value would be : $\Chi^2 =46.927$

And the rejection zone would be: $\chi^2 < 10.52 \cup \chi^2 46.927$

Since our calculated value is NOT in the rejection zone we FAIL to reject the null hypothesis.

Answer from: Quest

Ithink the answer will be d yes

Answer from: Quest

Hello sorry can’t good luck tho

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