Los Angeles Mission College



Chapter 10 Hypothesis Testing KEYHypothesis Testing: One Population MeansFor each problem below:a. Setup: Write the null and alternative hypothesisb. Find the p-value using Statcrunch. (Stat-t stats-one sample-with summary-sample mean ___, sample SD ___, sample size ___, hypothesis test ___, set inequality ___ -computec. Conclusion: Interpret the results in a complete sentence. d. Find the corresponding CI using Statcrunch (click on edit to go back and change to CI instead of hypothesis test)e. How does the CI support the conclusion in the hypothesis test?1. The manager at a local Starbucks wants to make sure that customers wait less than 4 minutes from the time they order to the time they pick up their coffee. In order to test this, twenty random customers were selected and the staff measured the number of minutes between when the person ordered and when their order was ready. The sample mean was 2.870 minutes and the sample standard deviation was 1.379 minutes. The histogram below shows the wait times of the twenty customers. A. Why does this data meet the assumptions necessary to perform a hypothesis test? The distribution of the histogram is roughly normal (symmetric).B. Use a 1% significance level to test the claim that the average wait time is less than 4 minutes.n = 20x=2.870s = 1.379α = 0.01Ho: μ=4 (The mean wait time is 4 minutes.)HA: μ<4 (The mean wait time is less than 4 minutes.)P value from Statcrunch: 0.0008 which is unusual and 0.0008 < 0.01. Reject the null hypothesisConclusion: There is enough evidence to support the claim that the average wait time is less than 4 minutes.C. Find the 98% CI (since significance level is 0.01 all in left tail). Then explain how it supports the conclusion from the hypothesis test.CI: (2.087 3.653) One can be 98% confident that the true mean wait time is between 2.1 and 3.7 minutes. Since population mean of 4 minutes is not in the CI, then there is evidence that the wait time is not 4 minutes. In fact, the CI shows that it is less. (Which is the same conclusion as above.)The manager can claim that the average wait time for their customers is less than 4 minutes.2. California is famous for its giant Redwood trees. Redwood trees are the tallest plants on Earth with a mean height of 240 feet. A biologist believes that the mean is actually greater. A random sample of 47 California Redwood trees was taken and their heights measure. The sample mean average height was 248 feet with a standard deviation of 26 feet.A. Why does this data meet the assumptions necessary to perform a hypothesis test? n ≥ 30B. Use a 5% significance level to test the claim that Redwood trees have an average height greater than 240 feet.n = 47x=248s = 26α = 0.05Ho: μ=240 (The mean height is 240 feet.)HA: μ>240 (The mean height is more than 240 feet.)P value from Statcrunch: 0.0202 which is unusual and 0.0202 < 0.05. Reject the null hypothesisConclusion: There is enough evidence to support the claim that Redwood trees have an average height greater than 240 feet.C. Find the 90% CI (since significance level is 0.05 all in the right tail). Then explain how it supports the conclusion from the hypothesis test.CI: (241.63, 254.37) One can be 90% confident that the true mean tree height is between 242 and 254 feet.Since 240 is not in the CI, there is evidence that the mean tree height is not 240 feet. In fact, the CI shows that it is greater. (Which is the same conclusion as above.)3. Mike wants to know the average price of a hamburger. He randomly selects 24 randomly selected restaurants and records the price of a regular hamburger. The sample mean price was $3.88 and the sample standard deviation was $1.14. A histogram of the data is below. A. Why does this data meet the assumptions necessary to perform a hypothesis test? The distribution of the histogram is roughly normal (symmetric).B. Use a 5% significance level to test the claim that the average price of a hamburger is greater than $3.50.n = 24x=$3.88s = $1.14α = 0.05Ho: μ=3.50 (The mean hamburger price is $3.50.)HA: μ>3.50 (The mean hamburger price is more than $3.50.)P value from Statcrunch: 0.058 which is not unusual and 0.058 is not less than <0.05. Cannot reject the null hypothesisConclusion: There is not enough evidence to support the claim that the average price is greater than $3.50.C. Find the 90% CI (since significance level is 0.05 all in the right tail). Then explain how it supports the conclusion from the hypothesis test.CI: (3.48, 4.28) One can be 90% confident that the true mean hamburger price is between $3.48 and $4.28.Since $3.50 is in the CI, there is not enough evidence that the mean hamburger price is not $3.50. Thus there is not enough evidence to support that the mean price is greater. (Which is the same conclusion as above.)Hypothesis Testing: ProportionsPerform a hypothesis test for each of the following claims at a significance level of 5% for all:For each problem below:a. Setup: Write the null and alternative hypothesisb. Find the p-value using Statcrunch. (Stat-proportion stats-one sample-with summary- successes ___, observations, Ho:p = ____, HA: p (inequality) ___ -computec. Conclusion: Interpret the results in a complete sentence. d. Find the corresponding CI using Statcrunch (click on edit to go back and change to CI instead of hypothesis test)e. How does the CI support the conclusion in the hypothesis test?1. Wikipedia claims that 10% of people are left handed. A sample of 250 randomly selected adults found that 32 of them were left handed. A. Are the conditions to perform a hypothesis test met? Use np0(1-po) ≥ 10 (Assume all other conditions have been met. 250(0.10)(0.90) 22.5 ≥10B. Is there enough evidence to claim that Wikipedia’s claim of 10% is wrong?p=32250=0.128Ho: p=0.10 (The percent of people that are left handed is 10 %.)HA: p≠0.10 (The percent of people that are left handed is not 10 %.)P value from Statcrunch: 0.14 which is not unusual and 0.14 is not less than <0.05. Cannot reject the null hypothesis.Conclusion: There is not enough evidence to support the Wikipedia’s claim that the percent of people that are left handed is not 10 %. C. Find the 95% CI (two tailed test). Then explain how it supports the conclusion from the hypothesis test.CI: (0.087, 0.169) One can be 95% confident that the true percent of people that are left handed is between 8.7% and 16.9% feet.Since 10% is in the CI, then there is not enough to show that the percent of left handers is no longer or different from 10%. Thus, there is not enough evidence to support Wikipedia’s claim that the percent of left handers is not 10%. (Which is the same conclusion as above.)2. According to a past CNN report, 93% of Americans also own a traditional phone (landline). Has the percentage since decreased as more Americans are opting to only use a cell phone? A random sample of 500 Americans was taken and 454 of them owned a traditional phone. A. Are the conditions to perform a hypothesis test met? Use np0(1-po) ≥ 10 (Assume all other conditions have been met. 500(0.93)(0.07) 32.55 ≥10 B. Test the claim that the percentage has decreased as more Americans are opting to only use a cell phone.p=454500=0.908Ho: p=0.93 (The percent of people that own a landline phone is 93 %.)HA: p<0.93 (The percent of people that own a landline is less than 93 %.)P value from Statcrunch: 0.0269 which is unusual and 0.0269 <0.05. Reject the null hypothesis.Conclusion: There is enough evidence to support the claim that the percent of Americans owning a landline is less than 93%. Therefore, the percentage has decreased as more Americans are opting to only use cell phones.C. Find the 90% CI (since significance level is 0.05 all in the left tail). Then explain how it supports the conclusion from the hypothesis test.CI: (0.887, 0.929) One can be 90% confident that the true percent of Americans owning a landline is between 88.7% and 92.9%.Since 93% is not in the CI, there is evidence that the percent of Americans owning a landline is not 93%. In fact, the CI shows that it is less. (Which is the same conclusion as above.) Note: In this case it is really close. If you round too much, you might not get the same conclusion.3. An analysist believes that more Americans are using credit cards today compared to the past. According to an article in USA Today, 74% of Americans have at least one credit card. However, this claim seems a little low. In order to verify the claim that more than 74% of Americans have a credit card, a random sample of 900 Americans was taken and 76% of them owned a credit card.A. Are the conditions to perform a hypothesis test met? Use np0(1-po) ≥ 10 (Assume all other conditions have been met. 900(0.74)(0.26) 173.16 ≥10B. Test the analysists’ claim.p=0.76For Statcrunch: We need the number of people who responded yes to having at least one credit card (#successes). x900=0.76-→solving for x=684Ho: p=0.74 (The percent of Americans that have at least one credit card is 74 %.)HA: p>0.74 (The percent of Americans that have at least one credit card is greater than 74 %.)P value from Statcrunch: 0.0857 which is not unusual and 0.0857 is not less than 0.05. Cannot reject the null hypothesis. Conclusion: There is not enough evidence to support the claim that more than 74% of Americans have at least one credit card.C. Find the 90% CI (since significance level is 0.05 all in the right tail). Then explain how it supports the conclusion from the hypothesis test.CI: (0.737, 0.783) One can be 90% confident that the true percent of Americans that have at least one credit card is between 73.7% and 78.3%.Since 74% is in the CI, then there is not enough to show that the percent of Americans having at least one credit card is no longer or different from 74%. Thus, there is not enough evidence to support that more than 74% of Americans have at least one credit card. (Which is the same conclusion as above.)4. The center for disease control estimates that 45.2 million people, or 19.3% of all adults (18 and older) in the U.S. smoke cigarettes. Cigarette smoking is the leading cause of preventable death in the U.S., accounting for approx. 443,000 deaths, or 1 of every 5 deaths each year. But is the claim that 19.3% really accurate? In a random sample of 3000 adults, 581 smoked. A. Are the conditions to perform a hypothesis test met? Use np0(1-po) ≥ 10 (Assume all other conditions have been met. 3000(0.193)(0.807) 467.25 ≥10B. Test the claim that the reported percent of 13.3% of all adults smoke is accurate.p=5813000=0.194Ho: p=0.193 (The percent of adults that smoke is 19.3 %.)HA: p ≠0.193 (The percent of adults that smoke is not 19.3 %.)P value from Statcrunch: 0.9263 which is not unusual and 0.9263 is not less than 0.05. Cannot reject the null hypothesis.Conclusion: There is not enough evidence to support the claim that the percent of adult smokers is not 19.3%.C. Find the 95% CI (two tailed test). Then explain how it supports the conclusion from the hypothesis test.CI: (0.180, 0.208) One can be 95% confident that the true percent of adult smokers is between 18.0% and 20.8%.Since 19.3% is in the CI, then there is not enough evidence to show that the percent of adult smokers is no longer or different from 19.3%. Thus, there is not enough evidence to support the claim that the percent of adult smokers is not 19.3%. (Which is the same conclusion as above.)5. A recent article by Sports Illustrated claims that within 5 years of retirement, 60% of NBA players are broke. The article also indicates that for NFL players, the percentage may be higher. In order to test the claim by Sports Illustrated that more than 60% of retired NFL players are broke, a random sample is taken. 200 retired NFL players are selected and 123 were found to have gone broke.A. Are the conditions to perform a hypothesis test met? Use np0(1-po) ≥ 10 (Assume all other conditions have been met. 600(0.60)(0.40) 48 ≥10B. Test the claim by Sports Illustrated that more than 60% of retired NFL players are broke.p=123200=0.615Ho: p=0.60 (The percent of retired NFL players that are broke is 60 %.)HA: p>0.60 (The percent of retired NFL players that are broke is greater than 60 %.)P value from Statcrunch: 0.3325 which is not unusual and 0.3325 is not less than 0.05. Cannot reject the null hypothesis.Conclusion: There is not enough evidence to support the claim that the percent of retired NFL players that are broke is more than 60%.C. Find the 90% CI (since significance level is 0.05 all in the right tail). Then explain how it supports the conclusion from the hypothesis test.CI: (0.558, 0.672) One can be 90% confident that the true percent of retired NFL players that are broke is is between 55.8% and 67.2%.Since 60% is in the CI, then there is not enough to show that the percent of retired NFL players that are broke is no longer or different from 60%. Thus, there is not enough evidence to support the claim that the percent of retired NFL players that are broke is more than 60%. (Which is the same conclusion as above.)6. The recession forced millions of Americans into homelessness according to estimates by The National Alliance to End Homelessness. A 2009 study reported that about 2% (1.5 million) of American children were homeless during this time. If the recession did force more families into homelessness, then subsequent studies should reflect this. A random sample of 10,000 children was taken in 2012 and found that 314 of them were homeless. If the recession did force more families into homelessness, then we would expect the percent of homeless children to have increased in 2012 from the aftermath of the recession.A. Are the conditions to perform a hypothesis test met? Use np0(1-po) ≥ 10 (Assume all other conditions have been met. 10000(0.02)(0.98) 196 ≥10B. Test the claim that the percent of homeless children increased due to the recession. p=31410000=0.0314Ho: p=0.02 (The percent of homeless children is 2 %.)HA: p>0.02 (The percent of homeless children is more than 2 %.)P value from Statcrunch: 0.0001 which is unusual and 0.0001 <0.05. Reject the null hypothesis.Conclusion: There is enough evidence to support the claim that the percent of homeless children is more than 2%. Therefore, there is evidence to suggest that the percent of homeless children increased due to the recession.C. Find the 90% CI (since significance level is 0.05 all in the right tail). Then explain how it supports the conclusion from the hypothesis test.CI: (0.029, 0.034) One can be 90% confident that the true percent of homeless children is between 2.9% and 3.4%.Since 2% is not in the CI, there is evidence that the percent of homeless children is not 2%. In fact, the CI shows that it is higher. There is enough evidence to support the claim that the percent of homeless children is more than 2%. (Which is the same conclusion as above.) 7. Childhood obesity has more than tripled in the past 30 years. The percentage of children aged 6-11 years in the U.S. who were obese increased from 7% in 1980 to nearly 20% in 2008. If this trend continues we can expect the percent of young children that are obese in 2012 to be significantly greater than 21%. In order to test this claim, a random sample of 800 children in the U.S. was taken and 179 were found to be obese.A. Are the conditions to perform a hypothesis test met? Use np0(1-po) ≥ 10 (Assume all other conditions have been met. 800(0.21)(0.79) 132.72 ≥10B. Test the claim that the percent of obese children in 2012 is significantly greater than 21.p=179800=0.224Ho: p=0.21 (The percent of obese children is 21 %.)HA: p>0.21 (The percent of obese children is greater than 21 %.)P value from Statcrunch: 0.1698 which is not unusual and 0.1698 is not less than 0.05. Cannot reject the null hypothesis.Conclusion: There is not enough evidence to support the claim that the percent of obese children is greater than 21%.C. Find the 90% CI (since significance level is 0.05 all in the right tail). Then explain how it supports the conclusion from the hypothesis test.CI: (0.200, 0.248) One can be 90% confident that the true percent of obese children is between 20.0% and 24.8%.Since 21% is in the CI, then there is not enough to show that the percent of obese children is no longer or different from 21%. Thus, there is not enough evidence to support the claim that the percent of obese children is greater than 21%. (Which is the same conclusion as above.)8. About 1 in every 3 U.S. adults (about 68 million) has high blood pressure which increases the risk for heart disease and stroke, a leading cause of death in the U.S. Another website claims that the true percentage is actually lower. To test this claim, 500 adults are randomly selected and 165 are found to have high blood pressure.A. Are the conditions to perform a hypothesis test met? Use np0(1-po) ≥ 10 (Assume all other conditions have been met. 500(0.33)(0.67) 110.55 ≥10B. Test the claim. p=165500=0.330 and p=13=0.333 (population proportion)Ho: p=0.333 (The percent of adults with high blood pressure is 33.3 %.)HA: p<0.333 (The percent of adults with high blood pressure is less than 33.3 %.)P value from Statcrunch: 0.4434 which is not unusual and 0.444 is not less than 0.05. Cannot reject the null hypothesis.Conclusion: There is not enough evidence to support the claim that the percent of adults with high blood pressure is less than 33.3%C. Find the 90% CI (since significance level is 0.05 all in the left tail). Then explain how it supports the conclusion from the hypothesis test.CI: (0.295, 0.365) One can be 90% confident that the true percent of adults with high blood pressure is between 29.5% and 36.5%.Since 33.3% is in the CI, then there is not enough to show that the percent of adults with high blood pressure is no longer or different from 33.3%. There is not enough evidence to support the claim that the percent of adults with high blood pressure is less than 33.3% (Which is the same conclusion as above.) ................
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