In December 2004, 51% of students in high school were satisfied with the lunch e

Question

In December 2004, 51% of students in high school were satisfied with the lunch es supplied through the school In May 2010, an organization conducted a pol of 912 students in high school and asked if they were satisfied with the lunches supplied through the school. Of the 912 surveyed, 410 indicated hey were satisfied Does this suggest the proportion of students satisfied with the quality of lunches has decreased? (a) What does it mean to make a Type Il error for this test? b) If the researcher decides to test this hypothesis at the = 0.10 level of significance, compute the probability of making a Type ll error, , if the true population proportion is 0.45. What is the power of the test? (c) Redo part (b) if the true population proportion is 0.49. (a) What does it mean to make a Type ll error for this test? Choose the correct answer below. O A. Ho is not rejected and the true proportion of high school students who are not satisfed with the quality of kunches is equal to 0 51 B. Ho s rejected and the true proportion of high school students who are not satisfied with the quality of lunches is less than 0 51. O C. Ho is not rejected and the true proportion of high school students who are not satisied with the qualiy f unches is less than 0.51 of making a Type II error, , if the true population b) If the research er decides test is hypothesis at the proportion is 0.45. What is the power of the test? 10 level of significance, compute the probability Power - Type integers or decimals rounded to four decimal places as needed.) (c) Redo part (b) if the true population proportion is 0,49 Power - Click to select your answer(s) hat

Explanation / Answer

let f denotes the number of students satisfied with the lunch out of 912 students

and p be the proportion of students who are satisfied with lunch

so f~Bin(912,p)

hence E[f/912]=p

hence f/912 is an unbiased estimator of p.

V[f/912]=912*p*(1-p)/9122=p(1-p)/912

since sample size=912 is very high

asymptotically, f/912~N(p,p(1-p)/912)

we want to test whether the proportion of students satisfied with lunch decreased from 51% or not.

so H0:p=0.51 vsa H1:p<0.51

now (f/912-p)/sqrt[p(1-p)/912] asymptotically follows a N(0,1)

so our test statistic is T= (f/912-0.51)/sqrt[0.51(1-0.51)/912] which under H0 follows N(0,1)

a) so type II error is accepting H0 when H0 is actually false.

now H0 is false implies the number of students satisfied with lunch is not less than 0.51

or, number of students not satisfied with lunch is less than 0.51

hence correct option is option c: H0 is not rejected and the true proportion of high school students who are not satisfied with the quality of lunches is less than 0.51

b) level of significance is alpha=0.10

since the alternate hypothesis is left tailed. so H0 is rejected iff t<-z0.1 where t is the observed value of T. z0.1 is the upper 0.1 point of N(0,1)

so type II error=beta=P[accepting H0| H0 is false]=P[T>-z0.1|p=0.45]

=P[(f/912-0.51)/sqrt[0.51(1-0.51)/912]>-z0.1|p=0.45]

=P[f/912>0.51-1.281552*sqrt[0.51*0.49/912]] where f/912~N(0.45,0.45*0.55/912)

=P[(f/912-0.45)/sqrt[0.45*0.55/912]>(0.488786-0.45)/sqrt[0.45*0.55/912]]

=P[Z>2.354426576] where Z~N(0,1)

=0.009275654 [answer]

hence power=1-beta=0.990724346 [answer]

c) if the treu proportion becomes 0.49

then beta=P[f/912>0.51-1.281552*sqrt[0.51*0.49/912]] where f/912~N(0.49,0.49*0.51/912)

=P[(f/912-0.49)/sqrt[0.49*0.51/912]>(0.488786-0.49)/sqrt[0.49*0.51/912]]

=P[Z>-0.073338661]

=0.5292317 [answer]

so power=1-0.5292317=0.4707683 [answer]

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