**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.

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.