Respuesta :
Answer:
[tex]\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54[/tex]
The degrees of freedom are:
[tex] df=n-1=10-1=9[/tex]
Now we can calculate the p value using the alternative hypothesis:
[tex]p_v =2*P(\chi^2 >13.54)=0.279[/tex]
Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces
Step-by-step explanation:
Assuming the following data:"12.14 12.05 12.27 11.89 12.06
12.14 12.05 12.38 11.92 12.14"
We can calculate the sample deviation with this formula:
[tex]s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
[tex]n=10[/tex] represent the sample size
[tex]\alpha=0.05[/tex] represent the confidence level
[tex]s^2 =0.0217 [/tex] represent the sample variance
[tex]\sigma^2_0 =0.12^2= 0.0144[/tex] represent the value to verify
Null and alternative hypothesis
We want to determine whether the standard deviation differs from 0.12 ounce, so the system of hypothesis would be:
Null Hypothesis: [tex]\sigma^2 = 0.0144[/tex]
Alternative hypothesis: [tex]\sigma^2 \neq 0.0144[/tex]
The statistic can be calculated like this;
[tex]\chi^2 =\frac{n-1}{\sigma^2_0} s^2[/tex]
[tex]\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54[/tex]
The degrees of freedom are:
[tex] df=n-1=10-1=9[/tex]
Now we can calculate the p value using the alternative hypothesis:
[tex]p_v =2*P(\chi^2 >13.54)=0.279[/tex]
Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces
