1, In this problem, we focus on running a hypothesis test when you have a summary of the data. Context: In 2014, Gallup conducted a
1, In this problem, we focus on running a hypothesis test when you have a summary of the data.
Context: In 2014, Gallup conducted a poll where 42% of those surveyed said they were "Independents" (as opposed to Democrat or Republican). You take a random sample of 74 people at your campus. You believe that the percentage of Independents on your campus is lower than the 42% found in the Gallup poll. You want to test that claim. In your sample, 27 students said they were Independents.
Your Null Hypothesis is:Ho:p=0.42
Your Alternative Hypothesis is:Ha:p<0.42
In this problem, the definition of "success" is someone who says they are Independent since we are testing a claim about the proportion of respondents who say they are Independent. The number of "observations" is the total number of people in your survey.
Run a hypothesis test for proportions for this situation. Be sure that you carefully enter the value of 0.42 for both the null proportion value and for the alternative proportion value. Also be sure that you have the inequality correctly set to <.
After you compute the test results, answer the following:
A) What is the sample proportion? This is labeled as "Sample Prop." in the output.
Answer: Sample proportion =.(Round to FOUR decimal places)
B) What is the test statistic? This is labeled as Z-Stat. Remember, this test is based on the assumption that the distribution is normal. We can make this assumption becausenpo=74(0.3649)>10 andn(1PO)=74(0.6351)>10. Recall that the test statistic tells us how many standard deviations from the mean this particular sample is.
Answer: Test-statistic, z=.(Round to FOURdecimal places)
Note that your sample proportion IS below the Gallup poll of 42%. However, is this result due to normal random variation that is built into samples? Or, is the sample so unlikely that you can attribute it to "something else." That is why we need the P-value...
C) What is the P-value for this test? This is the probability of getting a sample like the one you for which you have a summary.
Answer: P-value =(Round to FOURdecimal places)
From your results above, you see that the probability of getting a sample like the one you have is about 17%. That's pretty high. If we set ourlevel at 0.05, this is above that. We are more likely to get this kind of sample. So, we fail to reject the null hypothesisHo:p=0.42 (Remember, we either reject the null, or we fail to reject it. Those are the ONLY two options available to us.) Since we cannot reject the null (Ho:p=0.42), we do not have evidence that our campus hada smaller proportion of independents.
2, In this problem, we focus on running a hypothesis test when you have a summary of the data. In particular, we will look at a test involving a mean (quantitative variable) when we know or can assume we know, the population standard deviation. This is an important point. If we know, we can use the z-distribution. If we DO NOT knowwe will need to use a different distribution and test.
(Made up) Context:Let's suppose that the age of college students nationally is distributed normally, with a national average of 25 years and a national standard deviation of 2.6years.You believe that students at your college are older than the national average. You randomly choose22 random students from your college, and survey responses indicate an average age of 26 with a sample standard deviation of 2.9 years. Is there evidence that students are your college are older than the national average at the=0.05 level?
First, let's see if we can use a z-test. Since we have a quantitative variable, we first check if our sample size is large enough (n30?). This is NOT true, as we only have 22students in our sample. However, we know that age is normally distributed in the population. Therefore, we CAN use the z-test. (See this [LINK] for a handy chart on how to know when you can use the z-test.)So we proceed...
Your Null Hypothesis isHo:=25. We assume that our sample is no different than the national average and want to see if we have any evidence to reject that assumption.
Your Alternative Hypothesis is:Ha:>25
Once we open up the correcttool/screen, we need to be sure we enter the correct values into the boxes. For a one sample z-test with summary, we first enter the sample mean, which is 26.
The box labeled standard deviation is for the population standard deviation, which we are given. It is 2.6.(Notice that we are using thepopulationstandard deviation, not the sample standard deviation.)
The sample size here is 22 students.
We also need to make sure that the null and alternative hypotheses are correctly filled in. Enter 25 into each of those two boxes. Be sure to change the inequality symbol to > so it matches our alternative hypothesis.
Now run your test and look at the output to answer the following questions:
A) What is the test statistic? This is labeled as Z-Stat(since we are using the z-distribution).
Answer: Test-statistic, z=.(Round to FOURdecimal places)
Note that your sample meanIS above the national mean of 25.Is this result due to normal random variation that is in samples? Or, is the sample so unlikely you can attribute it to "something else." This is why we need the P-value.
B) What is the P-value for this test? This is the probability of getting a sample like the one for which you have a summary.
Answer: P-value =(Round to FOURdecimal places)
From your results above, you see that if we assume the null hypothesis is true, then the probability of getting a sample like the one you have is about 3% (0.03). That's pretty low. Since we set ourlevel at 0.05, this p-value is belowthat.As a result, we can reject the null and accept the alternative. There is evidence that students at our college are older than the national average. We say the results are "statistically significant (at the 0.05 level)."
C) What is the standard error? This is labeled as "Std. Err."
Answer: Standard Error=(Round to FOURdecimal places)
The standard error is the standard deviation of the sample means,nn. If you go back and check2.622 you would see this matches what StatCrunch gives for the Std. Err. StatCrunch uses this and the populationmean to find the P-value. Look at the following diagram and see if you can connect all the pieces together for this problem. If you can, you are one step closer to mastering the ideas in this topic and how they are connected to previous content in the course.
3, In this problem, we focus on running a hypothesis test when you have a summary of the data. In particular, we will look at a test involving a mean (quantitative variable) when we DO NOT know, the population standard deviation. Ifwe know, we can use the z-distribution. If we DO NOT know, we will need to use at-test.
(Made up) Context:Suppose in a survey of 46 students at your college, the mean monthly rent reported of the sample is $612, with sample standard deviation of $303.You have data showing that the mean monthly rent for college students in that entire city is $702.
Is there evidence that students at your college pay less rent than students in the city,at the=0.05 level?
that sincen30 and we do not know thepopulationstandard deviation, we need to use a t-test rather than a z-test.
Your Null Hypothesis is:Ho:=702 We assume that our sample is no different than the national average. We want to see if we have evidence to reject that assumption.
Your Alternative Hypothesis is:Ha:<702
Once we open up the correcttool/screen, we need to be sure we enter the correct values into the boxes. For a one sample t-test with summary, enter the sample mean, sample standard deviation, and the sample size. Be sure to change the hypothesis test values and inequality sign to match the null and alternative hypotheses.
Now run your test. Look at the output to answer the following questions:
A) What is thedegrees of freedom? This is labeled as DF. It is equal ton1
Answer: DF=.
B) What is the test statistic? This is labeled as the T-Stat
Answer: T-Stat =(Round to FOURdecimal places)
C) What is the P-value for this test? This is the probability of getting a sample like the one you for which you have a summary.
Answer: P-value =(Round to FOURdecimal places)
From your results above, you see that the probability of getting a sample like the one you have is between .02 and .03. This is belowourlevel at 0.05.This means we can reject the null and accept the alternative. Namely, there is evidence that students at our college pay less rent than those in the city overall. We say the results are "statistically significant (at the 0.05 level)."
4, Suppose you are testing the following claim: "More than 34% of workers got their job through networking." Express the null and alternative hypotheses in symbolic form for a hypothesis test (enter as a percentage).
H0:p(Enter percentages, not decimals. Example 99%, not 0.99)
Ha:p(Enter percentages, not decimals. Example 99%, not 0.99)
Use the following codes to enter the following symbols:
enter>=
>enter>
enter<=
enter!= 5, Suppose you are testing the following claim: "The mean weight of male nurses working at a local hospital is less than 190 lbs." Express the null and alternative hypotheses in symbolic form for a hypothesis test. H0: H1: Use the following codes to enter the following symbols: enter>= >enter> enter<= enter!= 6, You are conducting a study to see if the probability of a true negative on a test for a certain cancer is significantly less than 15%. WithHa: p<15% you obtain a test statistic ofz=2.096 Find the p-value accurate to 4 decimal places. p-value = 7, You are conducting a study to see if the probability of a true negative on a test for a certain cancer is significantly different from 0.81. Thus you are performing a two-tailed test. Your sample data produce the test statisticz=2.723. Find the p-value accurate to 4 decimal places. p-value = 8, You wish to test the following claim at a significance level of=0.002. Ho:p=0.25 Ha:p<0.25 You obtain a sample of sizen=518 in which there are 111 successful observations. a, What is the test statistic for this sample? test statistic =Round to 3 decimal places. b, What is the p-value for this sample? P-value =Use TechnologyRound to 4 decimal places. c, The p-value is... d, This test statistic leads to a decision to... e, As such, the final conclusion is that... 9, You are conducting a study to see if the probability of a true negative on a test for a certain cancer is significantly more than 0.23. You use a significance level of=0.001 H0:p=0.23 H1:p>0.23 You obtain a sample of sizen=631 in which there are 180 successes. What is the test statistic for this sample? test statistic =(Report answer accurate to 3 decimal places.) What is the p-value for this sample? p-value =(Report answer accurate to 4 decimal places.) The p-value is... This test statistic leads to a decision to... As such, the final conclusion is that... 10, You wish to test the following claim (HaHa) at a significance level of=0.01 Ho:=52.1 Ha:<52.1 You believe the population is normally distributed and you know the standard deviation is=6.6=6.6. You obtain a sample mean ofx=50.7 for a sample of sizen=32 What is the test statistic for this sample? test statistic =(Report answer accurate to 3 decimal places.) What is the p-value for this sample? p-value =Use Technology(Report answer accurate to 4 decimal places.) The p-value is... This test statistic leads to a decision to... As such, the final conclusion is that... 11, You wish to test the following claim (HaHa) at a significance level of=0.02 Ho:=66.3 Ha:<66.3 You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of sizen=10 with meanx=54.5 and a standard deviation ofs=13.3 What is the test statistic for this sample? test statistic =Round to 3 decimal places What is the p-value for this sample? p-value =Use TechnologyRound to 4 decimal places. The p-value is... This test statistic leads to a decision to... As such, the final conclusion is that... 12, You wish to test the following claim (HaHa) at a significance level of=0.001 Ho:=68.7 Ha:68.7 You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of sizen=111 with meanx=67.5 and a standard deviation ofs=13.9 What is the test statistic for this sample? test statistic =(Report answer accurate to 3 decimal places.) What is the p-value for this sample? p-value =(Report answer accurate to 4 decimal places.) The p-value is... This test statistic leads to a decision to... As such, the final conclusion is that... 13, You wish to test the following claim (Ha) at a significance level of=0.001 Ho:=85.9 Ha:<85.9 You believe the population is normally distributed, but you do not know the standard deviation. You obtain the following sample of data: Column A. B Column. C Column. D Column E Column 55.7 68.7 82.2 84.5 104.5 64.3 65.9 87.2 78 99.3 67.3 65.9 91. 73.4. 118.2 65.1. 107.3. 98.7. 55.7. 96.9 84.5. 66.6. 83.1. 85.4. 103.7 78.4. 87.7. 78. 94.1. 80.3 52.1. 126.3. 75.5. 100.7. 58.4 106.3. 41.7. 88.1. 98.1. 113.9 66.6. 102.9. 99.3. 77.5. 110.9 Note: To save vertical scrolling space, the data set is shown here with five columns. In Statcrunch, you will need to make sure all the data are in ONE column. Use Technology What is the test statistic for this sample? test statistic =(Report answer accurate to 4 decimal places.) What is the p-value for this sample? p-value =(Report answer accurate to 4 decimal places.) The p-value is... This test statistic leads to a decision to... As such, the final conclusion is that... 14, Erin in Philadelphia is conducting a study to see if the accuracy rate for fingerprint identification is significantly less than 0.53.Erin obtains a relevant sample of sizen=136 in which there are 58 people whose fingerprints are accurately identified. Perform an appropriate hypothesis test to help Erin make a statistically correct conclusion about this scenario. Use a significance level of=0.002 Perform a hypothesis test.
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Authors: Herbert Amann, Yoshikazu Giga, Hideo Kozono, Hisashi Okamoto, Masao Yamazaki
1st Edition
3034809395, 9783034809399
