The national center for education statistics monitors many aspects of elementary and secondary

Educated mothers The National Center for Education Statistics monitors many aspects of elementary and additional education nationwide. Their 1996 numbers are oft en supplied as a baseline to assess transforms. In 1996, 31% of students reported that their mothers had graduated from college. In 2000, responses from 8368 students uncovered that this figure had grown to 32%. Is this proof of a readjust in education and learning level among mothers?

a) Write appropriate hypotheses.

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b) Check the assumptions and also conditions.

c) Perform the test and uncover the P-value.

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d) State your conclusion.

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e) Do you think this difference is meaningful? Explain.


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Given: The National Center for Education Statistics monitors many type of aspects of elementary and also secondary education and learning nationwide. Their 1996 numbers are oft en used as a baseline to assess transforms. In 1996, 31% of students reported that their mothers had graduated from college. In 2000, responses from 8368 students discovered that this figure had grown to 32%. Is this proof of a change in education and learning level among mothers. To Find: a) Write proper hypotheses. b) Check the assumptions and problems. c) Perdevelop the test and uncover the P-value. d) State your conclusion. e) Do you think this distinction is meaningful? Exordinary. Solution: Let us assume that alpha = 0.05 Also, P0 = 0.31% , n = 8368 , p ^ = 0.32 a) The null hypothesis and the different hypothesis have the right to be characterized as follows H0 : P = 31 % (It state that populace propercentage is equal to the worth mentioned in the case.) Ha : P not equal to 31% b) We can use one proportion Z test given that we are all set in trial and error the case around one population propercent. Condition : (i)It is a random sample although assumed (ii) The provided...
value 8368 is less than 10% of all students. (iii) Success faitempt condition is likewise satisfied np and n(1-p) are both atleast 10. Here np = 8368 x ( 0.31) = 2594.08 >= 10 n( 1 - p) = 8368 ( 1 - 0.31) = 5773.92 > =10 c) The test statistics is z = ( p - p0) / root < p0 (1 - p0) / n> = = ( 0.32 - 0.31 ) / root < 0.31 ( 1 - 0.31 ) 8368 = 1.98 P = P < Z 1.98 > = 2 P ( Z Reject H0. Tright here is substantial evedience to assistance the claim. e) The difference is not systematic bereason the percent just differ by 1% and also this is a really tiny distinction.