Math 156 Sample Questions Exam 3
Math 156 review for exam 3 Chaps 11 - 15
1. Suppose I must pass through three stoplights on the way to school. If the first is red with probability .2 , the second red with probability .3 and the third red with probability .5, what is the chance I stop at most once on my way to school?
add the prob of stop not at all = .8*.7*.5
to the prob of stopping once = .2*.7*.5 + .8*.7*.5 + .8*.7*.5 (stop at first only second only and third ony)
What assumptions are you making? the lights are independent
2. Consider the probabilities listed below for the color of a randomly chosen car:
bluewhitegoldsilverother
.12.21.13.18.36
a) Consider the experiment of randomly selecting two cars. What is the probability that least one is gold?
1 - prob both are not gold = 1 - .87^2
b) If you randomly sample 10 cars and count the number that are gold, use to find the probability that this number is 2. 10 nCr 2 .13^2 .87^8
c) Use to find the approximate probability that in a sample of 100 cars, more than 10 are gold.
z = (10 - 13)/(100*.13*.87)^.5 = -/89 prob of more than = 1 - .1894
3. You forgot to study for your sociology quiz. It is an 8 question multiple choice quiz, with 4 possible answers for each question. You must score 6 or more correct to get a passing grade. If you just randomly guess:
a) What is the mean and standard deviation of the number of questions you get correct?
this is binomial so mean = n p = 8*.25 = 2 SD = (8*.25*.75)^.5 = 1.22
b) What is the probability of getting exactly 6 correct? 8 nCr 6 .25^6 .75^2 = .0038
c) What is the chance you pass the quiz? .0038 + .000366 + .0000153
4. Use the normal approximation to estimate the probability that in 600 rolls of a fair die, there would be more than 112 sixes.
normal approximation to binomial z = (112 - 100)/(600*(1/6)*(5/6))^.5 = 1.31 rob = .0951
5. Suppose a 95% confidence interval for the mean of a population is (17.5, 18.9).
a) What is the estimated mean from the data? 18.2
b) What is the margin of error? .7
c) Explain in plain language what the interval tells you. the population mean is likely to be in the interval
d) Would you expect around 95% of the sampled values to fall in this interval? Why or why not?
not necessarily. the interval tells where the population mean is, not sampled values. with large samples the interval will contain a very small percentage of sampled values
e) Suppose the sample size is 800. Under what conditions can you use the one sample z statistic: ?
with large samples such as this we need only SRS
6.a) Suppose a 95% confidence interval for the mean of a population is (99.3, 104.8).
Explain in plain language what the interval tells you. the population mean is likely to be in the interval
b) Suppose a 95% confidence interval for the mean is quoted to be (99.3, 104.8). Are sampled values likely to be in this interval? Why or why not? not necessarily. the interval tells where the population mean is, not sampled values. with large samples the interval will contain a very small percentage of sampled values
c) Use to produce a 95% confidence interval for the mean, if your sample consists of 25 items, whose mean is 47 and the assumed population standard deviation is 8. 47 ± 1.96 * 8/5
7. a) Produce a 99% confidence interval for the mean if your sample consists of 10 items, whose mean is 27.15 and the assumed population standard deviation is 1.25. 27.15 ± 2.576*1.25/3.1622
b) What must be true in order for the interval to have been constructed by a valid procedure? this is a small sample so we need a normal population distribution and SRS
8. a) Consider two confidence intervals for the mean of a population:
(6.1 , 13.9) and ( 4.84 , 15.15 ) . One is a 95% confidence interval and one is a 99% confidence interval. Which is which? Defend your answer. ( 4.84 , 15.15 ) is wider and so goes with the higher confidence level of 99%
b) Consider two confidence intervals for the mean of a population:
( 9.21 , 10.79) and ( 9.44 , 10.56 ) . One is from a sample of size 80 and one is from a sample of size 40 . Which is which? Defend your answer. ( 9.21 , 10.79) is wider and so goes with the smaller sample size of 40
9. a) Suppose we wish a 98% confidence interval to have a margin of error of .25, and the population standard deviation is assumed to be 3.25. Determine what sample size is required.
gives ( 2.326 * 3.25 /.25)^2 = 914.33 so the sample size 915
b) Suppose the standard error of the mean is .15 and the sample size is 200. About how big is the standard deviation? SEM = SD / root n so SD = SEM * root n = 2.121
c) In a recent study it was claimed that lengths of trout in Lake Pueblo have a mean greater than those upstream, which average 10 inches. State the null and alternative hypothesis in this setting.
null: = 10 alternative: > 10
d) Suppose a test results in a p-value of .0028. What can you say about the null hypothesis and alternative hypothesis? reject the null, there is strong evidence against the null in favor of the alternative, random variation would almost never produce results such as those observed
10. a) What happens to a confidence interval as the level of confidence is increased from 95% to 99%.
it gets wider
b) What happens to a confidence interval when the sample size increases from 10 to 160. it gets narrower
c) Suppose we wish a 95% confidence interval to have a margin of error of .09 when the population standard deviation is assumed to be 1.23. Use to determine what sample size is required.
n = (1.96 * 1.23 / .09 )^2 = 717.52 so the sample size is 718
d) Suppose the standard error of the mean is 11.5 and the sample size is 400. About how big is the standard deviation? 11.5 * 400^.5 = 230
11.a) A recent study claims that CSU - Pueblo students tend to average more than 15 credit hours per semester. State the null and alternative hypothesis in this setting. null: = 15 alternative: > 15
b) Suppose a test results in a p-value of .07. What can you say about the null hypothesis and alternative hypothesis? some evidence against the null hypothesis but not considered statistically significant. fail to reject the null hypothesis. random variation could sometimes produce results such as those observed.
12. a) Suppose z = - 2.03 and , . What is the p-value and what are your conclusions?
p-value is .0212 reject the null there is statistically significant evidence against the null in favor of the alternative. random variation rarely produces results such as those observed.
b) Is it possible for a result to be significant at the 1% level but not at the 5% level? Explain.
no if the p-value is less than ,01 it will be less than .05
c) If a two-sided test with fails to reject the null hypothesis at the 5% level, what can you say about the corresponding 95% confidence interval? 125 is in the interval
d) Suppose the null hypothesis is true. What is the chance the hypothesis test produces a result that is deemed statistically significant at the 5% level? .05
13.a) What size must z be, to be considered statistically significant at the .01 level if the alternative is two sided? 2.576 b) What size must z be, to be considered statistically significant at the .01 level if the alternative is one sided? 2.326
14.a) Is it possible for a result to be significant at the 1% level but not at the 5% level? Explain.
no if the p-value is less than ,01 it will be less than .05
b) If a two-sided test with rejects the null hypothesis at the 5% level, what can you say about the corresponding 95% confidence interval? 85 is not in the interval
c) Suppose you use z to perform an hypothesis test and n = 10. What needs to be true for your results to be valid? SRS normal population distribution
15.a) Suppose and z = 2.03. What are your conclusions? What is the p-value?
p-value .9788 fail to reject the null hypothesis there is no evidence against the null in favor of the alternative. random variation rarely produces results such as those observed, but the data indicates the true mean is probably greater than 15 not less than 15
b) Suppose and z = 2.03. What are your conclusions? What is the p-value?
p-value .0212 reject the null hypothesis there is statistically significant evidence against the null in favor of the alternative. random variation rarely produces results such as those observed.
c) Suppose and z = 2.03. What are your conclusions? What is the p-value?
p-value .0424 reject the null hypothesis there is statistically significant evidence against the null in favor of the alternative. random variation doesn't very often produce results such as those observed.
16. a) Perform a test of if , , and .
Quote an approximate p-value from the tables in your conclusions.
z =(5.75 - 5)/(2.5/60^.5) = 2.32 p-value = 2 * (1 - .9896) = .0208
reject the null hypothesis. there is statistically significant evidence against the null in favor of the alternative. random variation rarely produces results such as those observed.
Formulas Binomial mean standard deviation probability
z-score z-statistic
standard error confidence interval sample size