In may, the fill weights of 6-pound boxes of laundry soap had a mean of 6.13 pounds with a standard deviation of 0.095. the goal was to decrease the standard deviation. the company decided to adjust the filling machines and then test h0: σ = 0.095 against h1: σ < 0.095. in june, a random sample of size n = 20 yielded [(x)] = 6.10 and s = 0.065.1. what is the value of your test statistic? 2.what is the critical value for the test statistic at an α = 0.05 significance level? 3. what is the approximate p-value of the test?

1.

2.

3.

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".

represent the sample mean

n = 20 sample size

s= 0.065

the value that we want to test

represent the p value for the test

t represent the statistic

significance level

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is less than 0.095, so the system of hypothesis are:

H0:

H1:

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

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

1. What is the value of your test statistic?

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

2.What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a left tailed test the critical zone it's on the left tail of the distribution. On this case we need a quantile on the chi square distribution with 19 degrees of freedom that accumulates 0.05 of the area on the left tail and 0.95 on the right tail.

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.05,19)". And our critical value would be

Since our calculated value is less than the critical value we to reject the null hypothesis.

3. What is the approximate p-value of the test?

For this case since we have a left tailed test the p value is given by:

If we compare the p value and the significance level given we see that so then we have enough evidence to reject the null hypothesis that the true population deviation it's higher than 0.095 at 5% of significance.

D) $25

Step-by-step explanation:

good job figuring out on your own btw