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.

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, =0.05

By using chi square test χ^2 = ∑  =  

And another formula is χ^2 =

'S' be the sample standard deviation = 12.1

Given size of the sample 'n' = 22

Step 2:-

By using chi square test  

substitute all  given values

now chi square test =

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

the solution is 12x-6y

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.