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 .

Respuesta :

Answer:

[tex]\chi^2 =\frac{10-1}{18} 23.04 =11.52[/tex]

The degrees of freedom are:

[tex] df =n-1=10-1=9[/tex]

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

[tex]\chi^2_{\alpha/2}= 2.70[/tex]

[tex]\chi^2_{1-\alpha/2}= 19.02[/tex]

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

Step-by-step explanation:

Information given

[tex]n=10[/tex] represent the sample size

[tex]\alpha=0.05[/tex] represent the confidence level  

[tex]s^2 =4.8^2= 23.04 [/tex] represent the sample variance obtained

[tex]\sigma^2_0 =18[/tex] 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: [tex]\sigma^2 = 18[/tex]

Alternative hypothesis: [tex]\sigma^2 \neq 18[/tex]

The statistic would be given by:

[tex]\chi^2 =\frac{n-1}{\sigma^2_0} s^2[/tex]

And replacing we got:

[tex]\chi^2 =\frac{10-1}{18} 23.04 =11.52[/tex]

The degrees of freedom are:

[tex] df =n-1=10-1=9[/tex]

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

[tex]\chi^2_{\alpha/2}= 2.70[/tex]

[tex]\chi^2_{1-\alpha/2}= 19.02[/tex]

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

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