Hypothesis Testing of Single Mean and Single Proportion: Examples 1.8 2008/06/06 17:39:41 GMT-5 2008/08/20 11:09:36.391 GMT-5 Barbara Illowsky illowskybarbara@deanza.edu Susan Dean deansusan@deanza.edu Connexions cnx@cnx.org elementary statistics This module provides a examples of Hypothesis Testing of Single Mean and Single Proportion as a part of Collaborative Statistics collection (col10522) by Barbara Illowsky and Susan Dean. Jeffrey, as an eight-year old, established an average time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster by using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims. For the 15 swims, Jeffrey's average time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preconceived α=0.05. Setting up the Hypothesis Test:Since the problem is about a mean (average), this is a test of a single population mean.Ho: μ=16.43 Ha: μ16.43For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "" tells you this is left-tailed. Calculating the distribution needed:Random variable: X = the average time to swim the 25-yard freestyle. Distribution for the test: X is normal (population standard deviation is known: σ =0.8) X ~ N ( μ , σ X n ) Therefore, X ~ N ( 16.43 , 0.8 15 ) μ = 16.43 comes from H 0 and not the data. σ = 0.8 , and n = 15 . Calculate the p-value using the normal distribution for a mean:p-value=P(X -<16)= 0.0187 where the sample mean in the problem is given s 16.p-value=0.0187 (This is called the actual level of significance.) The p-value is the area to the left of the sample mean, 16.Graph:

μ=16.43 comes from Ho. Our assumption is μ=16.43.Interpretation of the p-value: If Ho is true, there is a 0.0187 probability (1.87%) that Jeffrey's mean (or average) time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is not happening randomly. It is a rare event. Compare α and the p-value: α = 0.05 p-value = 0.0187 α > p-value Make a decision: Since α>p-value, reject Ho.This means that you reject μ=16.43. In other words, you do not think Jeffrey swims the 25-yard freestyle in 16.43 seconds but faster with the new goggles.Conclusion: At the 5% significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds. The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Stats and press ENTER. Arrow down and enter 16.43 for μ0 (null hypothesis), .8 for σ, 16 for the sample mean, and 15 for n. Arrow down to μ: (alternate hypothesis) and arrow over to μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p=0.0187) but it also calculates the test statistic (z-score) for the sample mean. μ16.43 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with z=-2.08 (test statistic) and p=0.0187 (p-value). Make sure when you use Draw that no other equations are highlighted in Y= and the plots are turned off.When the calculator does a Z-Test, the Z-Test function finds the p-value by doing a normal probability calculation using the Central Limit Theorem (Chapter 7): P ( X 16 ) = 2nd DISTR normcdf ( -10 ^ 99 , 16 , 16.43 , 0.8 / 15 ) . The Type I and Type II errors for this problem are as follows:The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)The Type II error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.) Historical Note: The traditional way to compare the two probabilities, α and the p-value, is to compare their test statistics (z-scores). The calculated test statistic for the p-value is -2.15. You can find the test statistic for α=0.05 in the normal table. The z-score for an area to the left equal to 0.05 is midway between -1.65 and -1.64 (0.05 is midway between 0.0505 and 0.0495). The z-score is -1.645. Since -1.645>-2.15 (which demonstrates that α>p-value), reject Ho. Traditionally, the decision to reject or not reject was done in this way. Today, comparing the two probabilities α and the p-value is very common and advantageous. For this problem, the p-value, 0.0158 is significantly smaller than α, 0.05. You can be confident about your decision to reject. It is difficult to know that the p-value is significantly smaller than α by just examining the test statistics. The graph shows α, the p-value, and the two test statistics (z scores).

A college football coach thought that his players could bench press an average of 275 pounds. It is known that the standard deviation is 55 pounds. Three of his players thought that the average was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) 205(3) 215(3) 225(1) 241(2) 252(2) 265(2) 275(2) 313(2) 316(5) 338(2) 341(1) 345(2) 368(2) 385(1). (Source: data from Reuben Davis, Kraig Evans, and Scott Gunderson.)Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press average is more than 275 pounds. Setting up the Hypothesis Test:Since the problem is about a mean (average), this is a test of a single population mean. H o : μ = 275 H a : μ > 275 This is a right-tailed test. Calculating the distribution needed:Random variable: X = the average weight lifted by the football players.Distribution for the test: It is normal because σ is known. X ~ N ( 275 , 55 30 ) x = 286.2 pounds (from the data) n=30 σ=55 pounds (Always use σ if you know it.) We assume μ=275 pounds unless our data shows us otherwise. Calculate the p-value using the normal distribution for a mean:p-value=P( X >286.2)=0.1323 where the sample mean is calculated as 286.2 pounds from the data.Interpretation of the p-value: If Ho is true, then there is a 0.1323 probability (13.23%) that the football players can lift a mean (or average) weight of 286.2 pounds or more. Because a 13.23% chance is large enough, a mean weight lift of 286.2 pounds or more is happening randomly and is not a rare event.

Compare α and the p-value: α = 0.025 p-value = 0.1323 α p-value Make a decision: Since αp-value, do not reject Ho.Conclusion: At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds. The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:Put the data and frequencies into lists. Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Data and press ENTER. Arrow down and enter 275 for μ0, 55 for σ, the name of the list where you put the data, and the name of the list where you put the frequencies. Arrow down to μ: and arrow over to >μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p=0.1331, a little different from the above calculation - in it we used the sample mean rounded to one decimal place instead of the data) but it also calculates the test statistic (z-score) for the sample mean, the sample mean, and the sample standard deviation. μ>275 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with z=1.112 (test statistic) and p=0.1331 (p-value). Make sure when you use Draw that no other equations are highlighted in Y= and the plots are turned off. Statistics students believe that the average score on the first statistics test is 65. A statistics instructor thinks the average score is higher than 65. He samples ten statistics students and obtains the scores 65 65 70 67 66 63 63 68 72 71. He performs a hypothesis test using a 5% level of significance. The data are from a normal distribution. Setting up the Hypothesis Test: A 5% level of significance means that α=0.05. This is a test of a single population mean. H o : μ = 65 H a : μ > 65 Since the instructor thinks the average score is higher, use a ">". The ">" means the test is right-tailed. Calculating the distribution needed:Random variable: X = average score on the first statistics test. Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given. You are only given n=10 sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student-t.Use tdf. Therefore, the distribution for the test is t9 where n=10 and df=10-1=9. Calculate the p-value using the Student-t distribution:p-value=P( X >67)=0.0396 where the sample mean and sample standard deviation are calculated as 67 pounds and 3.1972 from the data.Interpretation of the p-value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 67 pounds or more.

Compare α and the p-value:Since α = .05 and p-value = 0.0396 . Therefore, α > p-value .Make a decision: Since α>p-value, reject Ho.This means you reject μ=65. In other words, you believe the average test score is more than 65.Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks. The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:Put the data into a list. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 65 for μ0, the name of the list where you put the data, and 1 for Freq:. Arrow down to μ: and arrow over to >μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p=0.0396) but it also calculates the test statistic (t-score) for the sample mean, the sample mean, and the sample standard deviation. μ>65 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with t=1.9781 (test statistic) and p=0.0396 (p-value). Make sure when you use Draw that no other equations are highlighted in Y= and the plots are turned off. Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is the same or different from 50%. Joon samples 100 first-time brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level of significance. Setting up the Hypothesis Test:The 1% level of significance means that α=0.01. This is a test of a single population proportion. H o : p = 0.50 H a : p ≠ 0.50 The words "is the same or different from" tell you this is a two-tailed test. Calculating the distribution needed:Random variable: P' = the percent of of first-time brides who are younger than their grooms. DistributionDistribution for the test: The problem contains no mention of an average. The information is given in terms of percentages. Use the distribution for P', the estimated proportion. P ' ~ N ( p , p ⋅ q n ) Therefore, P ' ~ N ( 0.5 , 5 ⋅ 5 100 ) where p=0.50, q=1-p=0.50, and n=100. Calculate the p-value using the normal distribution for proportions: p-value = P ( P ' 0.47 or P ' > 0.53 ) = 0.5485 where x=53, p'=xn = 53 100 = 0.53 Interpretation of the p-value: If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion p' is 0.53 or more OR 0.47 or less (see the graph below).

μ=p=0.50 comes from Ho, the null hypothesis.p'=0.53. Since the curve is symmetrical and the test is two-tailed, the p' for the left tail is equal to 0.50-0.03=0.47 where μ=p=0.50. (0.03 is the difference between 0.53 and 0.50.) Compare α and the p-value:Since α = 0.01 and p-value = 0.5485 . Therefore, α p-value .Make a decision: Since αp-value, you cannot reject Ho.Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50%. The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:Press STAT and arrow over to TESTS. Press 5:1-PropZTest. Enter .5 for p0 and 100 for n. Arrow down to Prop and arrow to not equals p0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator calculates the p-value (p=0.5485) and the test statistic (z-score). Prop not equals .5 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with z=0.6 (test statistic) and p=0.5485 (p-value). Make sure when you use Draw that no other equations are highlighted in Y= and the plots are turned off. The Type I and Type II errors are as follows: The Type I error is to conclude that the proportion of first-time brides that are younger than their grooms is 50% when, in fact, the proportion is actually different from 50%. (Reject the null hypothesis when the null htpothesis is true). The Type II error is to conclude that the proportion of first-time brides that are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%. (Do not reject the null hypothesis when the null hypothesis is false.) Suppose the proportion of households that have three telephone numbers is 30%. The telephone company has reason to believe that the proportion of households is less than 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three telephone numbers. Setting up the Hypothesis Test: H o : p = 0.30 H a : p 0.30 Calculating the distribution needed:The random variable is P' = proportion of households that have three telephone numbers.The distribution for the hypothesis test is P ' ~ N ( 0.30 , 0.30 ⋅ 0.70 150 ) Calculate the p-value using the normal distribution for proportions: p' = 43150 Calculate the p-value using the normal distribution for proportions:The value that helps determine the p-value is p'. Calculate p' below.p'=xnwhere x = the number of successes What is a success for this problem? A success is having three telephone numbers in your household x=43, n= 150, and p'= ________ using the formula above. p' = 43150 Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately. Compare α and the p-value:Fill in the blanks. p-value = _________ p-value = 0.3594 Make a decision. _____________(Reject/Do not reject) H0 because____________. Assuming that α = 0.5, α < p-value. Do not reject H0 because there is not sufficient evidence that the proportions of households that have three telephones is 30% Write a conclusion. The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter p. The distribution for the test is normal. The estimated proportion p' is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived α=0.01, for comparison, and a 95% confidence interval computation. The poem is clever and humorous, so please enjoy it!Notice the solution sheet that has the solution. Look in the Table of Contents for the topic "Solution Sheets." Use copies of the appropriate solution sheet for homework problems. My dog has so many fleas, They do not come off with ease. As for shampoo, I have tried many types Even one called Bubble Hype, Which only killed 25% of the fleas, Unfortunately I was not pleased. I've used all kinds of soap, Until I had give up hope Until one day I saw An ad that put me in awe. A shampoo used for dogs Called GOOD ENOUGH to Clean a Hog Guaranteed to kill more fleas. I gave Fido a bath And after doing the math His number of fleas Started dropping by 3's! With his old shampoo I counted 42. At the end of his bath, I redid the math And the new shampoo had killed 17 fleas. So no I was pleased. Now it is time for you to have some fun With the level of significance being .01, You must help me figure out Use the new shampoo or go without? Setting up the Hypothesis Test:Ho: p=0.25 Ha: p>0.25 Calculating the distribution needed:In words, CLEARLY state what your random variable X or P' represents.P’ = The proportion of fleas that are killed by the new shampooState the distribution to use for the test.Normal: N ( 0.25 , ( 0.25 ) ( 1 - 0.25 ) 42 ) Test Statistic: t=2.3163 Calculate the p-value using the normal distribution for proportions:p-value=0.0103In 1 – 2 complete sentences, explain what the p-value means for this problem.If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 (1742) or more.Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the p-value.

Compare α and the p-value:Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using COMPLETE SENTENCES. alpha decision reason for decision 0.01 Do not reject Ho αp-value Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of flea that are killed by the new shampoo is more than 25%.Construct a 95% Confidence Interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the Confidence Interval.

Confidence Interval: (0.26,0.55) We are 95% confident that the true population proportion p of fleas that are killed by the new shampoo is between 26% and 55%.This is a weak test since the p-value is very close to alpha. In reality, one would probably do more tests. Central Limit Theorem Given a random variable (RV) with known mean μ and known variance σ 2 size 12{ {} rSup { size 8{2} } } {}, we are sampling with size n and we are interested in two new RV - sample mean, Xˉ size 12{ { bar {X}}} {},and sample sum,Σ X size 12{X} {}. If the size n of the sample is sufficiently large, then Xˉ size 12{ { bar {X}}} {}∼ N nμ σ 2 n and ΣX size 12{X} {} ∼ N nμ n σ 2 . In words, if the size n of the sample is sufficiently large, then the distribution of the sample means and the distribution of the sample sums will approximate a normal distribution regardless of the shape of the population. And even more, the mean of the sampling distribution will equal the population mean and mean of sampling sums will equal n times the population mean. The standard deviation of the distribution of the sample means, σ n , is called standard error of the mean. Standard Deviation A number that is equal to the square root of the variance and measures how far data values are from their mean. Notations: s for sample standard deviation and σfor population standard deviation.