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An overview of confidence intervals and hypothesis testing in statistics. It covers the mathematical concepts behind confidence intervals, the calculation of confidence intervals for different scenarios, and the logic of hypothesis tests. Topics include sampling for means with known and unknown population deviations, sampling for differences of means, and sampling for proportions. The document also includes examples and procedures for various statistical tests such as one-sample z test, one-sample t test, matched pairs t test, two-sample t test, one proportion z test, and two proportion z test.
Typology: Study notes
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1.1. Computing confidence intervals.
(1) Identify the desired confidence level C%. (2) Let O be observed value (of the sample). (3) Identify the standard deviation of the relevant (sampling) distribution, call it S. (4) Identify the kind of distribution which is relevant for the problem, z if the standard deviation of the entire population is known or t if you only know the deviation of your sample. (5) Look up the critical value associated to the confidence level C%. Call this critical value v∗. (6) The margin of error for the observation is v∗^ × S. In each setting one can work backwards from this equation to determine how large a sample is needed or what level of confidence is needed to result in a desired margin of error. (7) The confidence interval says that with C% probability the value of interest is between O − (v∗^ × S) and O + (v∗^ × S).
1.2. Instances of confidence intervals.
s 12 n 1 +^
s 22 n 2 , where the first sample has size^ n^1 and deviation^ s^1 and the second sample has size n 2 and deviation s 2. 1
p¯(1−¯p) n , where ¯p^ is the proportion assuming four more responses, two of which are positive. (This is “plus four” method, needed almost always).
pˆ 2 (1 − pˆ 2 ) n 2
2.1. Computing hypothesis tests.
(1) Identify the null and alternative hypotheses. (2) Let O be observed value (of the sample). (3) Identify the standard deviation of the relevant (sampling) distribution, call it S. (4) Identify the kind of distribution which is relevant for the problem, z if the standard deviation of the entire population is known or t if you only know the deviation of your sample. (5) Compute the standardized variable t or z = O−SH , which measures how many deviations away O is from the value H predicted by the null hypothesis. (6) If the distribution is normal, use Table A to calculate the probability of observing a value less than a negative z or bigger than a positive z. (7) If a t-statistic is needed, work “backwards” from Table C to find the probability of observing a value less than a negative t (by first converting a negative to a positive) or greater than a positive t. (8) This probability is equal to the probability of observing something less than/ bigger than O assuming the null hypothesis. It is called the P -value.
(9) If the P -value is less than α, we say the null-hypothesis has been rejected (and thus the alternative hypothesis accepted) at level α. If the P -value is not less than our desired α we say the test is inconclusive.
2.2. Instances of hypothesis testing.
s 12 n 1 +^
s 22 n 2 , where the first sample has size^ n^1 and deviation^ s^1 and the second sample has size n 2 and deviation s 2.
p¯(1−¯p) n , where ¯p^ is the proportion assuming four more responses, two of which are positive. (This is “plus four” method, needed almost always).
p ˆ(1 − pˆ)
1 n 1 +^
1 n 2
, where ˆp is the total proportion is positive responses over both samples.
3.1. Problem 1. In the following situations, determine which one of the following statistical procedures is appropriate to analyze the described data. Fill in each blank with the letter of the correct procedure. Procedures: A. one sample z test B. one sample t test C. matched pairs t test D. two sample t test E. one proportion z test
F. two proportion z test Situations: (1) To determine if a new restaurant might be economically viable, a pollster conducts a poll to find out how many people eat out more than once per week. (2) You wish to measure the effectiveness of a placebo treatment. You take volunteers for your experiment, assign half to take a cold-prevention medication (which they don’t know is a placebo) and the other half to do nothing. After 6 months you find out the proportion of each group which says it has had a cold in the last 6 months. (3) A researcher wishes to measure the effect of alcohol on reaction time. She measures the reaction times of 100 subjects with no alcohol, and the same subjects after two drinks. (4) In order to estimate the mean high school GPA of Oregonians, you take a random sample of 500 high school graduates and obtain their GPAs.
3.2. Problem 2. A petroleum company is testing a new blend of gasoline to determine if it results in better gas mileage. They test with 20 identical new automobiles. Each automobile gets 10 gallons of gas and is then driven until it runs out of gasoline. Half of the cars get the new mixture and half get the old mixture. The cars receiving the old mixture went for a mean distance of 278 miles on the ten gallons of gas, with standard deviation 2.09 miles. The cars receiving the new mixture went for a mean distance of 284 miles on the tank of gas, with standard deviation 3.1 miles. (a) Construct a 95% confidence interval for the difference between the mean mileage of cars receiving the new mixture and cars receiving the old mixture. (b) Is there significant evidence at the 0.01 level that the mean miles driven with the new mixtures is larger than the mean miles driven with the old mixture?
3.3. Problem 3. A random sample of 634 Washington state registered voters showed that 437 of them voted in the 2004 presidential election. A random sample of 714 Oregon registered voters showed that 535 of them voted in the same election. You are interested in the effect of voting by mail, and you wish to determine if this is evidence that more Oregonians vote than Washingtonians.
(1) State null and alternative hypotheses as full sentences. (2) Calculate your test statistic. (3) Calculate your P -value. (4) State your conclusion.
3.4. Problem 4. 46 mice were trained to run a maze. After the training period they were divided into two groups of 22 and 24 each. The first group was given water laced with caffeine, and the other was given ordinary water. After one week of this, the rats were timed in the mazes again. Group n Mean maze time Std. Dev. Caffeine 22 178 49 No caffeine 24 164 26 (1) Give a 95% confidence interval for the mean time in the maze for the caffeine treated mice. (2) Give a 95% confidence interval for the mean difference of the time for the caffeine treated mice minus the time for the non-caffeine treated mice.