CFA Practice Question

There are 434 practice questions for this study session.

CFA Practice Question

The supporters of Judge Hyde want to prove that she has the support of more than 60% of the voters in the district. To accomplish this, they take a random sample of 80 voters and find that 53 support Judge Hyde. For a hypothesis test at the 5% level of significance, the conclusion would be ______

A. Judge Hyde has more than 60% of the voters' support.
B. Judge Hyde has 66.25% of the voters' support.
C. there is no evidence that Judge Hyde has more than 60% of the voters' support.
Correct Answer: C

User Contributed Comments 18

User Comment
kaliokale how did he get the test value? what formular is that?
johans how did they derive the standard deviation?
yliu bernoli distribution, variance = p * (1 - p)
ehc0791 for sample proportion,
the test value = z value (p - p-sub0)/s
s = sqrt(p*(1-p)/n)
tabulator yliu, don't confuse people. ehc0791 - well done.
StanleyMo When the answer is too big, more than 4, we conclude that we do not have the evidence.
chamad Why using bernouli trial in this case.
labine bernoulli trial applies because you're dealing with the probability of sucess or failure(Hyde does or does not have voters' support). yliu is correct: variance for binomial distributions=p*(1-p), therefore the standard deviation is [p*(1-p)]^1/2 and the standard error of the mean is:
[p*(1-p)]^1/2/(n^1/2)=[p*(1-p)/n]^1/2=[0.6(1-0.6)/80]^1/2
FayeMulvaney additionally, remember the rule that if the p-value is smaller than or equal to the level of significance (alpha) one generally rejects the null hypothesis. In this case, p is greater therefore we do not reject it.

Could somebody please tell me how to calculate the critical value out of interest?
KimActuary labine: to calculate the sample s.d., why not use .6625? Isn't that the p for the sample? I don't understand why you would use .6 when that is a test p.
afficionado I thought the Bernomial distribution variance is:
n*p(1-p)instead of p(1-p). Can someone comment on this please?
johntan1979 There's been more than one similar question testing this formula before this in previous topics, so if you didn't get them back then, you will not get it now.

Test value = (p'-p)/sqrt p(1-p)/n
gill15 Did you guys skip all the other sections in this reading to not know where this comes from
Thediceman I get correct answer with using wrong P value... same question as KimActuary. why use .60 instead of .6625

thank you
sgossett86 THEDICEMAN & KIMACTUARY

I keep having the same problem on these questions.. using sample as P ...
Shaan23 I wouldnt use z test here. It specifically says in the notes if the population variance is UNKNOWN, use t score irrespective of sample size.

So even though we have n=80 which is approaching the Z table values --- it isnt the same number.
farrahkame Could we not use p = 1 - (table value of z-score)
AKA table value of z-score = 1- 0.6625 = 0.3375
look it up in the table to find the closest z-score being -0.42 (table value 0.3372)
and -0.42 would fall in the -1.645 < test value < 1.645
hence, fail to reject H0...

does that make sense?
SandraZake Guys isn't the critical value -1.645, confirmed by the previous question. Since the test value is 1.14 (positive) this is greater than the critical value and therefore the Ho is rejected. Let me know your thoughts.
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