Term
| You find that a set of data has a P-Value of 0.025, you conclude that the data is _________ at level alpha=0.05 but not at _________. |
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Definition
-Significant
- Any number < .05 |
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Term
| The critical value, Zstar, used for constructing a 97% confidence interval for a population mean is________? |
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Definition
(97+1)/2 = .985
Look .985 on Table A and .985=2.17 |
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Term
| The critical value, tstar, used for constructing a 96% confidence interval with degrees of freedom 17 for population mean is __________? |
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Definition
| Given 96% Con Int, and 17 Degrees of Freedom, look them up on Table C. The answer is 2.235 |
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Term
| Suppose we have a population distribution that is normal with mean = 1123 and a standard deviation of 150. IF we take a random sample of 140 individuals from the population and find the average xbar, then xbar follows the distribution of_______? |
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Definition
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Term
| The standard deviation of the standard Normal Distribution is________? |
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Definition
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Term
| In a Normal distribution, ____________ percent of the data falls below the median. |
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Definition
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Term
| If the scores on a particular test in a massive online lecture of 30,000 students have a Normal distribution with mean 74 and standard deviation of 10, then approximately 68% of the scores fall between ______________ and ________. IF you take a simple random sample of 193 students in the class, the probability that xbar is greater than 76 is_________? |
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Definition
Part one:
64 = 74 - 10 = xbar - 1*sigma
84 = 74 + 10 = xbar + 1*sigma
Part 2:
Since xbar is greater than 76, that means that the area to the right of 76 on the normal curve. Find z = (x-mu)/(s/sqrt(n))
(76-74)/(10/sqrt(193))=2.7785 (round 2.78) Look at Table A for 2.78, then you find 0.9973. Then the area to the right of this z-score is 1-.9973=.0027 |
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Term
| Bobby took the test and recieved a score of 87. Using the information from the previous problem, his score was ___________ standard deviations above the mean. |
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Definition
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Term
| Suppose that the correlation between the explanatory variable x and the response variable y is given by 0.52. If you switch the explanatory and response variables the correlation becomes____________??? |
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Definition
| .52 (stays the same, idk why) |
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Term
| In a random study of college students regarding marijuana use, 13 out of 34 male respondents admitted to occasionally smoking marijuana while 10 out of 35 females respondents admitted to occasionally smoking marijuana. In this situation you want to use a __________ confidence interval. If you did a significance test to see if the proportions for each group were different, the standard error used to compute the z statistic would be_________. |
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Definition
Part one:
- Plus 4
Part 2:
- 0.111
Two Sample Proportions:
- phatM (men)= (13+1)/(34+2)= .3889
- phatF (females)= .297
Now find the standard error of z statistic:
- Formula is (sqrt the entire formula) SQRT(phatM(1-phatm)/(nM+2)+phatF(1-phatF)/(nF+2)
-Plug numbers in
sqrt(.3889(1-.3889)/(34+2)+0.297(1-0.297)/(35+2) = .111 |
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Term
| Suppose you suspect that the average birthweight of golden retrievers is greater than the average birthweight of yellow labradors. You take an SRS of 24 newborn golden retrievers and find the mean weight to be 1.2 pounds with a standard deviation of .35 pounds. You then take an SRS of 21 yellow lab newborns and find the mean weight to be 1.1 lbs with a standard deviation of .4 lbs. If you do a significance test, the P-Value will lie between________ & _________. |
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Definition
| Answer: .15 & .2
Work:
Two sample T-Test. We have mean weight of both dogs: muG = golden retrievers
muY = Yellow Labs
Our alternative hypothesis is Ha: muG>muY. This means we are looking for a one sided P-Value.
First, We have the formula:
T= muG-muY/sqrt(stdG^2/nG + stdY^2/nY
1.2-1.1/sqrt(0.35^2)/24 + 0.4^2/21 = .88655
Go to table C, look at the line for 20 degrees of freedom (always go with the smaller one) and Find the numbers that bound .88665 which should be .86 and 1.064. Follow these two numbers downward on Table C till you get to the One Sided P-Values Column. You should get 1.5 |
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Term
| You want to estimate the proportion (p) of Defazio voters with a 95% confidence and a MofE no greater than 3%. You can accomplish this if your sample size is greater than or equal to ____________? |
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Definition
Answer: 1068
Work:
We have the formula N = phat(1-phat)/(m/zstar)2. If they dont give a sample for problems like these, we use 0.5 as phat.
Formula:
0.5(1-0.5)/(.03/1.96)^2 = 1067.11 (round up to 1068) |
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Term
| Suppose that in fact 8% of all adult men preferred Pepsi over Coke. Pepsi is concerned about this information and interviews a random sample of 1256 people from this population. In many such samples, the proportion phat of the 1256 people in the sample who prefer Pepsi over Coke would var according to (approximately) the distribution of ___________? |
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Definition
Formula: (0.08, sqrt(0.08(1-0.08)/1256)
N (phat, sqrt(phat(1-phat)/n) |
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Term
| The critical value, zstar, used for constructing a 93% confidence interval for population mean is ______? |
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Definition
Answer: 1.81 or 1.82
Steps: Find the number half way between 93 and 100: 96.5... Find the closest number to .965 inside table A = .9649. Find it's corresponding Zscore. which is 1.81 |
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Term
| Suppose GRE scores are approximately normal with a mean 1000 and a standard deviation of 80. What do you need to get to be in the 95th percentile?? |
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Definition
Answer: 1131.6
Steps: 95th percentile corresponds to 1.645. Thus, 1.645 = z = x-mu/sigma.
To find x = 1000 + 1.645(80)= 1131.2 |
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Term
| Suppose ACT scores are approximately normal with mean 19 and a standard deviation 2. Approximately ________ percent of ACT takers will score at least 24? |
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Definition
Answer: .62
Steps: The same as finding the area to the right. The z score is 24-19/2=2.5.
Find 2.5 on Table A (.9938)
1-.9938 = .0062 |
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Term
| A sprinter runs eight 100 yard sprints during practice. The sample space for counting the number of sprints are under 11 seconds is_________________? |
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Definition
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