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Basic Question 5 of 8
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 ______
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.
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.
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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Learning Outcome Statements
construct hypothesis tests and determine their statistical significance, the associated Type I and Type II errors, and power of the test given a significance level
CFA® 2025 Level I Curriculum, Volume 1, Module 8.