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Midterm Practice
Topic A
Topic B
Topic C
Topic D
100
The average GPA at a particular school is 2.89 with a standard deviation .63. A random sample of 25 students is collected. Use the Central Limit Theorem to approximate the mean and standard error of your sample.
The average is µ(ẍ) =2.89 standard error is ơ(ẍ)= ơ/√n =.63/√25 =.126
100
Imagine you are working as a research consultant for a large org. The org is interested in examining employee satisfaction. The org employs over 10,000 employees so having everyone complete the scale is not practical. The org decides to take a random sample of employees and only administer the scale to these employees. The org assumes the sample will represent the attitudes of the whole population of employees. The org is unsure as to how many people to administer the survey. One researcher at the org wants to use a sample of 20 employees whereas another wants to use a sample of 100. Which should they choose and why? Present statistical evidence for your answer.
We know that the formula for calculating the standard error of the mean includes n as the denominator. We can tell from this formula that as n increases, the standard error of the mean decreases. Since a smaller standard error of the mean indicates lower error and greater precision, a sample size of 100 would be a better choice to represent the overall employee satisfaction.
100
Describe the concept and importance of: Mutually Exclusive Exhaustive Independent Events
Exhaustive- set of events includes all possible outcomes ME- if the occurrence of one event precludes the occurrence of the other IE- when the occurrence or nonoccurrence of one has no effect on the occurrence or nonoccurrence of the other.
100
How can you increase power?
1) Increase the distance between the means 2)Increase sample size 3)Decrease st. dev of distribution 4) change alpha level, increase .01 to .05
100
Researchers examined performance of 28 students who answered multiple choice items on the SAT without having read the passages to which the items referred. The mean score (out of 100) was 46.6, with a standard deviation of 6.8. Random guessing would have been expected to result in 20 correct answers. A) Were these students responding at better than chance levels? B) If performance statistically better than chance, does it mean that the SAT test is not a valid predictor of future college performance?
A) t=20.70 on 27df we can reject the null B) this does not mean that the SAT is not a valid measure, but it does show that people who do well at guessing at answers also do well on the SAT. This is not very surprising.
200
A set of 5000 scores on a college readiness exam are known to e approximately normally distributed with a mean of 72 and standard deviation of 6. To the nearest integer value, how many scores are there between 63 and 75?
N=5000, µ=72, ơ = 6 Find the corresponding z values for x=63 and x=75. We get z=-1.5 and .5 62% of scores fall in the range .62*5000= approx. 3123 scores
200
Consider the following results of 10 tosses of a coin: H;T;T;T;T;H;T;H;T;T a) Estimate the probability of head (H) for this coin. b) Estimate the standard error of your estimate.
.3 and 0.14
200
To assess the accuracy of a laboratory scale, a standard weight that is known to weigh 1 gram is repeatedly weighed 4 times. The resulting measurements (in grams) are: 0.95, 1.02, 1.01, 0.98. Assume that the weighing by the scale when the true weight is 1 gram are normally distributed with mean. a)Use these data to compute a 95% confidence interval for. b) Do these data give evidence at 5% significance level that the scale is not accurate? Answer this question by performing an appropriate test of hypothesis.
Extra prob. 7
200
Assume that we want to test a null hypothesis about a single mean at alpha=.05, one tailed. Further assume all necessary assumptions met. Could there be a case in which we would be more likely to reject a true Ho than to reject a false one? (In other words, can power ever be less than alpha?)
Discuss and decide on the answer
200
Researchers reported an intervention program for women with abusive partners. The study involved a 10-week intervention program and a three year followup, and used an experimental (intervention) and control group. At the end of the 10wk intervention period the mean quality of life score for the intervention group was 5.03 with a standard deviation of 1.01 and a sample size of 135. For the control group the mean was 4.61 with standard deviation of 1.13 and a sample size of 130. Do these data indicate that the intervention was successful in terms of effect size for that difference?
t= 2.545, p<.05, which tells us that the quality of life was better for the intervention group.
300
A manufacturer of TV sets found that for the sets he produces, the lengths of time until the first repair can be described using a normal model with a mean of 4.5 years and a standard deviation of 1.5 years. If the manufacturer sets the warrantee so that only 10.2% of the 1st repairs are covered by the warrantee, how long should the warrantee last?
Find the zvalue that corresponds to p=.102 z=-1.27 Z=(x-µ)/ơ x=2.595 so approx. 2.6 years
300
A case of 10 men were given a special diet. It was desired to test weight loss in points at the end of a two week period. Hypothesize test given the following numbers. ∑đ column =-40 (d-đ)^2 =60
S(d)= √((∑(d-đ)^2)/(n-1)) =√60/10-1 =2.58lbs t=(đ-Do)/S(d)/√n = -4/(2.58/√10) = -4.88 df=10-1 =9 tcrit= -1.833 [-4.88] > [-1.833] therefore reject Ho
300
In 2000 the state of Vermont legislature approved a bill authorizing civil unions between gay or lesbian partners. This was a very contentious debate with very serious issues raised by both sides. How the vote split along gender lines may tell us something important about the different ways that males and females looked at this issue. What would you conclude from these data? Vote: YES NO Total Women 35 9 44 Men 60 41 101 Total: 95 50 145
X(squared)= 5.50, Reject Ho and conclude that women voted differently from men. Women were much more likely to vote for Civil Unions- the odds ratio is (35/9)/(60/41) = 3.89/1.46= 2.66, meaning women had 2.66 times the odds of voting for civil unions than men. That is a substantial difference and likely reflects fundamental attitude differences.
300
Explain the effect of power using BEAN Describe the relations between power and effects of: Beta error, effect size, alpha error, sample size
As beta error increases, power decreases As effect size increases, power increases (if no change to alpha or sample size) As alpha error increases, power increases (if no effect size or sample size change) As sample size increases, power increases (if no change in effect size or alpha)
300
Researchers compared the performance on SAT items of a group of 17 students who were answering questions about a passage after having read the passage with the performance of a group 28 students who had not seen the passage. The mean for the first group was 69.9 and standard deviation for the first group 10.6, whereas for the second group they were 46.6 and 6.8. A)What is the null hypothesis? B) What is the alt hypothesis? C) Run the appropriate t test D) Interpret the results *hint pool variances
A) null- there is not a significant different in test scores between those who have read the passage and those who have not. B) Alt- there is a significant difference between the two conditions C) t=8.89 on 43df if we pool the variances, difference significant D) We can conclude the students do better on this test if they read the passage on which they are going to answer questions.
400
Suppose that Bob can decide to go to work by one of three modes of transportation (care, bus, commuter train). Because of high traffic, he decides to go by car, there is a 50%chance he will be late. If he goes by bus, which has reserved lanes but can be overcrowded, the probability of being late is only 20%. The commuter train is almost never late, with a prob. of 1%. but is more expensive than the bus. Suppose Bob's coworker knows that Bob almost always takes the train to work, never takes the bus, but sometimes (10%) takes the car. What is the coworker's estimate of the prob. that Bob drove to work that day given that he was late?
P(L/C)=.5 P(L/B)= .2 P(L/T)= .01 P(B)= 0 P(C)= .1 P(T)= .9 Use Bayes Theorem => P(C/L)= .847
400
The operations manager of a large production plant would like to estimate the mean amount of time a worker takes to assemble a new electronic component. Assume that the standard deviation of this assembly time is 3.6 minutes. After observing 120 workers assembling similar devices, the manager noticed that the average time was 16.2 min. construct a 92% confidence interval for the mean assembly time. Estimate how many workers should be involved in this study in order to have the mean assembly time estimated up to +/- 15sec with 92% confidence.
A) [15.5, 16.8] B) 636
400
Assuming the population standard deviation = 3, how large should a sample be to estimate the population mean with a margin of error not exceeding 0.5?
We need a sample of size at least 139. Extra prob #2
400
A PhD candidate has impression that he must find significant results if he wants to defend his dissertation successfully. He wants to show a difference in social awareness, as measured by his own scale, between a normal group and a group of ex-delinquents. He has a problem, he has data to suggest that the normal group has a true mean of 38 and he has 50 of those subjects. He has access to 100 high school graduates who have been classed as delinquent in the past. Or, he has access to 25 high school dropouts who have a history of delinquency. He suspects that the high school graduates come from a population with a mean of approximately 35, whereas the dropout group comes from a pop with a mean of approx. 30. He can use only one of these groups. Which should he use? *Hint: ơ can be any value as long as same for both calculations
He should use the dropout group. Assuming equal standard deviations, the HS dropout group of 25 would result in a higher value of δ and therefore higher power. Calculate δ for both situation.
400
Researchers reported on several different workout plans for obesity. There were 29 participants in a martial arts condition, and they were weighed before and after treatment. The handout has the results for each participant. Did the girls in this group lose a statistically significant amount of weight? Calculate a 99% confidence limit on these results. Compute an effect size measure for these results.
Same as homework problem with different Ho and H1
500
Random Variable X has following distribution X 20 21 22 23 24 P(X=x) .2 .3 .2 .1 .2 Find P(X≤22) Find P(X>21) Find P(21≤X<24)
P(X≤22) = P(x=22)+P(x=21)+P(x=20)=.2+.3+.2=.7 P(X>21) = P(x=22)+P(x=23)+P(x=24)=.2+.1+.2=.5 P(21≤X<24)= P(x=21)+P(x=22)+P(x=23)=.3+.2+.1=.6
500
Suppose a consumer advocacy group would like to conduct a survey to find the proportion p of consumers who bought the newest generation of an MP3 player were happy with their purchase. a) How large a sample n should they take to estimate p with 2% margin of error and 90% confidence?
Extra prob. #5
500
Suppose we asked a group of participants whether they liked Monday Night Football, then made them watch a game and asked them again. Our interest lies in whether watching a game changes people's opinions. Out of 80 participants, 20 changed their opinion from Favorable to Unfavorable, while 5 changed from Unfavorable to Favorable. (The others did not change). Did watching the game have a systematic effect on opinion change? A) Run the test B) Explain how this tests the null hypothesis C) In this situation the test does not answer our question of whether watching football has a serious effect on opinion change. Why not?
A) X(squared)= 9.0 B) If watching monday night football really changes people's opinions (in a negative direction), then of those people who change, more should change from positive to negative than vise versa, which is what happened. C) The analysis does not take into account all of those people who did not change. It only reflects direction of change if a person changes.
500
If ơ=15, n=25 and we are testing Ho: µ=100 versus H1: µ>100, what value of the mean under H1 would result in power being equal to the probability of a Type II error? *Hint: try sketching two distributions, which areas are you trying to equate?
When µ=104.935, power will equal β
500
The handout has data on endorphin levels as a function of stress. Based on this data, what effect does increased stress have on endorphin levels? Calculate an effect size for the data
Same as problem 7.16 answer