an Intemet survey was e-mailed to 6997 subjects randomly selected from an online

Question

an Intemet survey was e-mailed to 6997 subjects randomly selected from an online group involved Use a 0.01 significance level to test the claim that the return rate is less than 20%. Use the P-value method and use the with ears. There were 1341 surveys returned. normal distribution as an approximation to the binomial distribution. dentify the null hypothesis and alternative hypothesis OA. Ho: p=0.2 Hy: p 0.2 O c. Ho p0.2 Hy: p= 0.2 Ho:p=0.2 H1:p#0.2 OB, Ho:p#02 H1 : p=0.2 OD, Ho: p = 0.2 H: p>0.2 O F Hoip> 0.2 H1:p=0.2 E. The test statistic is z Round to two decimal places as needed.) The P-value is Round to three decimal places as needed) Because the P-value is is less than 20%. thesgrif anonieve, henulhypothesis. There. w/ ovidence to support the claim that tho return rate Click to select your answerfs).

Explanation / Answer

The statistical software output for this problem is:

One sample proportion summary hypothesis test:
p : Proportion of successes
H0 : p = 0.2
HA : p < 0.2

option a is correct

sample propotion p^ = x/n =1341/6997=0.191

claimed proportion P=20% =0.20

alpha =0.01

STANDARD DEVIATION = SQRT(P(1-P) /N) = SQRT(0.2(1-0.2)/6997)= 0.00478

test stastic =( p^-0.2) / std = (0.191-0.2) /0.00478 = -1.78

test stastic =-1.78

since this is lowered tailed test p value = p( z< zobserved ) = P(Z<-1.78) =0.037

reject h0 if pvalue <alpha

decision : since p- value >=alpha we fail to reject the null hypothesis

ther is no sufficient evidence to conclude that p is less than 0.2

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