Question

The sugar content of the syrup in canned peaches is normally distributed. Suppose that the variance is thought to be σ2=18 (milligrams)2. A random sample of n=10 cans yields a sample standard deviation of s=4.8 milligrams. Part 1 (a) Test the hypothesis H0:σ2=18 versus H1:σ2≠18 using α=0.05 Find χ02 .

Answer 1

`\chi^2 =(10-1)/(18) 23.04 =11.52`

The degrees of freedom are:

`df =n-1=10-1=9`

Now we can calculate the critical value taking in count the alternative hypotheis we have two values:

`\chi^2_(\alpha/2)= 2.70`

`\chi^2_(1-\alpha/2)= 19.02`

Since the calculated value is between the two critical values we FAIL to reject the null hypothesis and we can't conclude that the true variance is different from 18

Explanation:

Information given

`n=10` represent the sample size

`\alpha=0.05` represent the confidence level

`s^2 =4.8^2= 23.04` represent the sample variance obtained

`\sigma^2_0 =18` represent the value to verify

System of hypothesis

We want to verify if the true variance is different from 18, so the system of hypothesis would be:

Null Hypothesis:
`\sigma^2 = 18`

Alternative hypothesis:
`\sigma^2 \\eq 18`

The statistic would be given by:

`\chi^2 =(n-1)/(\sigma^2_0) s^2`

And replacing we got:

`\chi^2 =(10-1)/(18) 23.04 =11.52`

The degrees of freedom are:

`df =n-1=10-1=9`

Now we can calculate the critical value taking in count the alternative hypotheis we have two values:

`\chi^2_(\alpha/2)= 2.70`

`\chi^2_(1-\alpha/2)= 19.02`

Since the calculated value is between the two critical values we FAIL to reject the null hypothesis and we can't conclude that the true variance is different from 18