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Hypothesis Testing, Exercises of Statistics

There are two types of hypotheses: The null hypothesis, , is the current belief. The alternative hypothesis, , is your belief; it is what you want to show.

Typology: Exercises

2022/2023

Uploaded on 03/01/2023

danmarino
danmarino 🇺🇸

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Download Hypothesis Testing and more Exercises Statistics in PDF only on Docsity! Lecture 10 Hypothesis Testing A hypothesis is a conjecture about the distribution of some random variables. For example, a claim about the value of a parameter of the statistical model. There are two types of hypotheses: The null hypothesis, , is the current belief. The alternative hypothesis, , is your belief; it is what you want to show. Examples: Each of the following situations requires a significance test about a population mean. State the appropriate null hypothesis and alternative hypothesis in each case. (a) The mean area of the several thousand apartments in a new development is advertised to be 1250 square feet. A tenant group thinks that the apartments are smaller than advertised. They hire an engineer to measure a sample of apartments to test their suspicion. (b) Larry's car consume on average 32 miles per gallon on the highway. He now switches to a new motor oil that is advertised as increasing gas mileage. After driving 3000 highway miles with the new oil, he wants to determine if his gas mileage actually has increased. (c) The diameter of a spindle in a small motor is supposed to be 5 millimeters. If the spindle is either too small or too large, the motor will not perform properly. The manufacturer measures the diameter in a sample of motors to determine whether the mean diameter has moved away from the target. Example: An air freight company wishes to test whether or not the mean weight of parcels shipped on a particular root exceeds 10 pounds. A random sample of 49 shipping orders was examined and found to have average weight of 11 pounds. Assume that the standard deviation of the weights is 2.8 pounds. Solution: Graphical Representation Suppose we want to test a set of hypotheses concerning a parameter based on a random sample . vs ̂ is the estimate of our parameter . Rejection Region (RR) is the specified values of the test statistics for which we reject . The probability that defines the critical region is called the size of the test or level of the significance of the test and is denoted by α. Example: The hourly wages in a particular industry are normally distributed with mean $13.20 and standard deviation $2.50. A company employs 40 workers paying them an average of $12.20 per hour. Can this company be accused of paying substandard wages? Use . Solution: Example: 85% of the general public is right-handed. A survey of 300 chief executive officers of large corporations found that 95% were right-handed. Is this difference in percentages statistically significant? Use . Find the P-value for the test. Solution: Tests for a Population Mean ( is known) where is the specified value of . eweeys SITIES : : — = = a Zz TEST FOR A POPULATION MEAN. a ye €R To test the hypothesis Ho: 1 = “o based on an SRS of size n from a S ovag fe lation with unknown mean yu and known standard deviation 7, compute the test statistic ye Sn X — po ~ ofa In terms of a standard Normal random variable Z, the P-value for a test of Ho against Hoi kh > [to 18 PE = Z) Aa: w < 6 is PZ =z) He: w # uo is 2P(Z > [al) Aa izI These P-values are exact if the population distribution is Normal and are approximately correct for large m in other cases. / CLT Recall: (one-sample t CI) Example: Founded in 1998, Telephia provides a wide variety of information on cellular phone use. In 2006, Telaphia reported that, on average, United Kingdom (U.K.) subscribers with third- generation technology (3G) phones spent an average of 8.3 hours per month listening to full-track music on their cell phones. Suppose we want to determine a 95% CI for the U.S. average and draw the following random sample of size 8 from the U.S. population of 3G subscribers: 5 6 0 4 11 9 2 3 The sample mean is ̅ and the standard deviation s = 3.63 with degrees of freedom n - 1 = 7. THE ONE-SAMPLE t TEST Suppose that an SRS of size n is drawn from a population having un- known mean u. To test the hypothesis Ho: 1 = “o based on an SRS of size n, compute the one-sample t statistic : Pee tl s/Jn In terms of a random variable T having the t(m — 1) distribution, the P- value for a test of Hp against Ha: > Mo is P(T = t) Hg: uw < to is PT <0) Hq: # Mo is 2P(T = It!) al These P-values are exact if the population distribution is Normal and are approximately correct for large n in other cases. Example: Suppose that, for the U.S. data in example before we want to test whether the U.S. average is different from the reported U.K. average. How to increase the power? * Increase a. A 5% test of significance will have a greater chance of rejecting the alternative than a 1% test because the strength of evidence required for rejection is less. * Consider a particular alternative that is farther away from suo. Values of that are in H, but lie close to the hypothesized value jp are harder to detect (lower power) than values of yz that are far from po. * Increase the sample size. More data will provide more information about ¥ so we have a better chance of distinguishing values of . * Decrease o. This has the same effect as increasing the sample size: more in- formation about uw. Improving the measurement process and restricting at- tention to a subpopulation are two common ways to decrease o. Back to Error Probabilities Example: The mean outer diameter of a skateboard bearing is supposed to be 22.000 millimeters (mm). The outer diameters vary Normally with standard deviation mm. When a lot of bearings arrives, the skateboard manufacturer takes an SRS of 5 bearings from the lot and measures their outer diameters. The manufacturer rejects the bearings if the sample mean diameter is significantly different from 22 at the 5% significance level. Suppose the producer and the manufacturer agree that a lot of bearings with mean 0.015 mm away from 22 should be rejected. Significance and Type I error: The significance level of any fixed level test is the probability of a Type I error. That is, is the probability that the test will reject when is in fact true. Power and Type II error: The power of a fixed level test to detect a particular alternative is 1 minus the probability of a Type II error for that alternative. Here we know and , which is quite rare. So in general, there are two ways to compare the means of two normal populations. This is due to the fact that there are two distinct possibilities: 1. and are unknown and equal. 2. and are unknown and unequal. Comparing Two Mean: Variances Unequal Assume and are unknown. We estimate them by and . Example: An educator believes that new directed reading activities in the classroom will help elementary school pupils improve some aspects of their reading ability. She arranges for a third-grade class of 21 students to take part in these activities for an eight-week period. A control classroom of 23 third-graders follows the same curriculum without the activities. At the end of the eight weeks, all students are given a Degree of Reading Power (DRP) test, which measures the aspects of reading ability that the treatment is designed to improve. The data appear in the table below: Comparing Two Means: Variances Equal (Pooled Test) Suppose we have two Normal populations with the same variances: , is unknown. The pooled two-sample t procedures: Choose an SRS of size from a Normal population with unknown mean and an independent SRS of size from another Normal population with unknown mean . A ( ) CI for is given by ( ̅ ̅ ) √ where is the value for density curve with area between and . To test the hypothesis , compute the pooled two- sample t statistic ̅ ̅ √ In terms of a random variable T having the distribution, the P-value for a test of against Example: Does increasing the amount of calcium in our diet reduce blood pressure? Examination of a large sample of people revealed a relationship between calcium intake and blood pressure, but such observational studies do not establish causation. A randomized comparative experiment gave one group of 10 people a calcium supplement for 12 weeks. The control group of 11 people received a placebo that appeared identical. Table below gives the seated systolic blood pressure for all subjects at the beginning and end of 12-week period, in millimeters of mercury. The table also shows the decrease of each subject. An increase appears as a negative entry.
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