According to a report issued by the CDC in 2013. 66 % of women 40 years of age a

Question

According to a report issued by the CDC in 2013. 66 % of women 40 years of age and over had a mammogram within the last two years. An oncologist at the CDC would like to know if there has been a change in this percentage over the last two years. A random sample of 649 women (40 and older) is taken, and each is asked the date of their last mammogran. . Verify that the assumptions have been satisfied to conduct a hypothesis test. Random sample? Select an answer The variable recorded was: Select an answer and n(1 - p)- Both of the values are greater t than 10: Select an answer It was found that 420 women in the sample had a mammogram in the last two years. Test the appropriate hypotheses using a significance level of 0.01 . 110 : Select an answer "v| | : select an answerT? ect an answer decision rule: reject Hn if probability Note: round the z-score to two decimal places cary at least four decimal places throughout all of your caleulations) Note: round the probability to four decimal places) Test Statistic: z probability Decision: Select an answer . Conclusion: At the 0.01 level, there Select an answer significant evidence to conelude the percentage of women 40 or older who have had a mammogram in the last two years is Select an answer 66.8%

Explanation / Answer

Verify the assumptions
Random sample? Yes, it is a random sample
The variable recorded was? no. of women 40 years of age and over had a mammogram within last 2 years
np = 649 * 0.668 = 433.532
np(1-p) = 649 * (1-0.668) = 215.468
both of teh values are greater than 10? Yes, true it is

HYPOTHESIS
Given that,
possibile chances (x)=420
sample size(n)=649
success rate ( p )= x/n = 0.6471
success probability,( po )=0.668
failure probability,( qo) = 0.332
null, Ho:p=0.668  
alternate, H1: p!=0.668
level of significance, alpha = 0.01
from standard normal table, two tailed z alpha/2 =2.576
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.64715-0.668/(sqrt(0.221776)/649)
zo =-1.1279
| zo | =1.1279
critical value
the value of |z alpha| at los 0.01% is 2.576
we got |zo| =1.128 & | z alpha | =2.576
make decision
hence value of |zo | < | z alpha | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.12793 ) = 0.25935
hence value of p0.01 < 0.2593,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.668, alternate, H1: p!=0.668
alpha = 0.01, reject Ho if < alpha
test statistic: -1.1279
critical value: -2.576 , 2.576
decision: do not reject Ho
p-value: 0.25935
at the 0.01 level, thers is suffcient evidence that it is not changed from 66.8%

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