Suppose that on a 100-point test, the goal is for the students' exam scores to have a standard deviation of less than 10 pts. A recent sample of 22 exams has a standard deviation of 12.1. Use alpha = 0.05.

H0: population variance > or = 100
H1: population variance < 100

What is the critical Chi-square value?



For the same hypothesis test, what is the Chi-Square statistic?



For the same hypothesis test, what is the p-Value?



For the same hypothesis test, interpret the results.

Respuesta :

Answer:

a) chi square test value = 32.2

b) The tabulated value 32.671 at α =0.05 level of significance

c) conclusion:-

null hypothesis is accepted

population variance is greater than or equal to 100

Step-by-step explanation:

Step1:-

Given size of the sample 'n' = 22 and standard deviation 'S'= 12.1

H0: population variance > or = 100

H1: population variance < 100

level of significance,[tex]\alpha[/tex] =0.05

By using chi square test χ^2 = ∑ [tex]\frac{(x-x^{-}) ^{2} }{sigma^2}[/tex]  =  

And another formula is χ^2 = [tex]\frac{nS^2}{variance}[/tex]

'S' be the sample standard deviation = 12.1

Given size of the sample 'n' = 22

Step 2:-

By using chi square test  [tex]χ^2 =\frac{nS^2}{б^2}[/tex]

substitute all  given values

now chi square test = [tex]\frac{22(12.1)^{2} }{100} = \frac{3221.02}{100} = 32.2[/tex]

calculated chi square test = 32.2

The degrees of freedom = n-1 in chi-square distribution

The degrees of freedom = 22-1 =21  at 5% level of significance

The tabulated value from chi square table = 32.671

calculated chi square test = 32.2< The tabulated value 32.671

so the null hypothesis is accepted

Therefore  population variance is greater than or equal to 100

Conclusion

a) chi square test value = 32.2

b) The tabulated value 32.671 at α =0.05 level of significance

c) conclusion:-

null hypothesis is accepted

population variance is greater than or equal to 100