Question

Consider the following two questions designed to assess quantitative literacy. 1. What is 15% of 1000? 2. A store is offering a 15% off sale on all TVs. The most popular television is normally priced at $1000. How much money would a customer save on the television during this sale? Suppose the first question is asked of 200 randomly selected college students, with 167 answering correctly; the second one is asked of a different random sample of 200 college students, resulting in 144 correct responses. Carry out a test of hypotheses at significance level 0.05 to decide if the true proportion of correct responses to the question without context exceeds that for the one with context. (Use P, for the true proportion students who answered the question without context correctly and P2 for the true proportion of students who answered the question with context correctly.) in USE SALT State the relevant hypotheses. O Ho: P.-P2 - 0 Ho: P.-P20 OH: P.-P2 > 0 HP - P2 - 0 Ho: P2 - P2 = 0 H.: P.-P2 > 0 O Ho: P2 - P2 < 0 4: P1 P2 = 0 OH: P1-P2 - 0 HP - P2*0 Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) Z 3.2458 X P-value = 0.0006 X

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Answer 1

Solution-1:

Ho:p1-p2=0

Ha:p1-p2>0

alpha=0.05

test statistic

z=p1^-p2^/sqrt(pbar*(1-pbar)*(1/n1+1/n2))

p1^=x1/n1=167/200= 0.835

p2^=x2/n2=144/200= 0.72

pbar=x1+x2/n1+n2=(167+144)/(200+200)= 0.7775

z=(0.835-0.72)/sqrt(0.7775*(1-0.7775)*(1/200+1/200))

z= 2.76492

z=2.76

p value in excel

==NORM.S.DIST( -2.76492,TRUE)

=0.002846838

P value=0.0028

ANSWER:

z=2,76

p-value=0.0028

Solution-2:

df ANOVA Source of Variation Between Groups Within Groups Sum of squares Mean squares F 2 0.199 0.099 24 1.163 0.048 2.049 Total 26 1.362

ANSWER:

F=2.05