For the situation described below, state the null hypothesis,H 0 ,and the alternative hypothesis,H a ,to be tested. (Enter != for≠as needed.)A statistical test is designed to show that the proportion...

For the situation described below, state the null hypothesis,H₀,and the alternative hypothesis,H_a,to be tested. (Enter != for≠as needed.)A statistical test is designed to show that the proportionpof defectives has decreased below0.9%.H
₀:H

_a
:

2.

[–/1 Points]DETAILSMENDSTAT15 9.1.005.0/6 Submissions UsedMY NOTESASK YOUR TEACHERFor the situation described below, state the null hypothesis,H₀,and the alternative hypothesis,H_a,to be tested. (Enter != for≠as needed.)A statistical test is designed to show that the meanμis different from200.H
₀:H

_a
:

3.

[–/1 Points]DETAILSMENDSTAT15 9.1.006.0/6 Submissions UsedMY NOTESASK YOUR TEACHERFor the situation described below, state the null hypothesis,H₀,and the alternative hypothesis,H_a,to be tested. (Enter != for≠as needed.)A researcher claims that a binomial proportionpis at most0.7. A statistical test is designed to disprove the researcher's claim.H
₀:H

_a
:

4.

[–/1 Points]DETAILSMENDSTAT15 9.1.011.0/6 Submissions UsedMY NOTESASK YOUR TEACHERA new variety of pearl millet is expected to provide an increased yield over the variety presently in use which is about90bushelsper acre. The new variety of millet produced an average yield ofx=98bushelsper acre with a standard deviation ofs=12.8bushelsbased on 40 one-acre yields.Find the value of the test statistic for testing the hypotheses that the new variety will increase yield. (Round your answer to two decimal places.)Is the value of the test statistic likely, assumingH₀is true?Since the test statistic is less than one standard deviation from the mean, it is a highly likely event, assumingH
₀is true.Since the test statistic is between one and two standard deviations from the mean, it is a somewhat likely event, assumingH
₀is true.Since the test statistic is between two and three standard deviations from the mean, it is a somewhat unlikely event, assumingH
₀is true.Since the test statistic is more than three standard deviations from the mean, it is a highly unlikely event, assumingH
₀is true.

5.

[–/1 Points]DETAILSMENDSTAT15 9.2.006.NVA0/6 Submissions UsedMY NOTESASK YOUR TEACHERConsider the following situation.a right-tailed test withα= 0.1Find the appropriate rejection regions. (Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused region.)z>zState your conclusion if the observed test statistic wasz=2.17.If appropriate, provide a measure of reliability for your conclusion.The null hypothesis is not rejected.The null hypothesis is rejected at the 10% level.The null hypothesis is rejected at the 0.1% level.You may need to use the appropriateappendix tableortechnologyto answer this question.

6.

[–/1 Points]DETAILSMENDSTAT15 9.2.007.NVA0/6 Submissions UsedMY NOTESASK YOUR TEACHERConsider the following situation.a two-tailed test at theα=0.1Find the appropriate rejection regions. (Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused region.)z>zState your conclusion if the observed test statistic wasz=2.19.If appropriate, provide a measure of reliability for your conclusion.The null hypothesis is not rejected.The null hypothesis is rejected at the 0.1% level.The null hypothesis is rejected at the 10% level.You may need to use the appropriateappendix tableortechnologyto answer this question.

7.

[–/1 Points]DETAILSMENDSTAT15 9.2.008.NVA0/6 Submissions UsedMY NOTESASK YOUR TEACHERConsider the following situation.a left-tailed test withα= 0.01Find the appropriate rejection regions. (Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused region.)z>zState your conclusion if the observed test statistic wasz=−2.42.If appropriate, provide a measure of reliability for your conclusion.The null hypothesis will be rejected at the 0.01% level.We do not reject the null hypothesis.The null hypothesis will be rejected at the 1% level.You may need to use the appropriateappendix tableortechnologyto answer this question.

8.

[–/1 Points]DETAILSMENDSTAT15 9.2.011.NVA0/6 Submissions UsedMY NOTESASK YOUR TEACHERFind thep-value for thez-test. (Round your answer to four decimal places.)a right-tailed test with observedz=1.14p-value =Determine the significance of the results.The results are highly significant.The results are statistically significant.The results are only tending toward statistical significance.The results are not statistically significant.You may need to use the appropriateappendix tableortechnologyto answer this question.

9.

[–/1 Points]DETAILSMENDSTAT15 9.2.012.NVA0/6 Submissions UsedMY NOTESASK YOUR TEACHERFind thep-value for thez-test. (Round your answer to four decimal places.)a two-tailed test with observedz=−2.72p-value =Determine the significance of the results.The results are highly significant.The results are statistically significant.The results are only tending toward statistical significance.The results are not statistically significant.You may need to use the appropriateappendix tableortechnologyto answer this question.

10.

[–/1 Points]DETAILSMENDSTAT15 9.2.013.NVA0/6 Submissions UsedMY NOTESASK YOUR TEACHERFind thep-value for thez-test. (Round your answer to four decimal places.)a left-tailed test with observedz=−1.86p-value =Determine the significance of the results.The results are highly significant.The results are statistically significant.The results are only tending toward statistical significance.The results are not statistically significant.You may need to use the appropriateappendix tableortechnologyto answer this question.

11.

[–/1 Points]DETAILSMENDSTAT15 9.2.016.0/6 Submissions UsedMY NOTESASK YOUR TEACHERA random sample ofn=50observations from a quantitative population produced a meanx=2.6and a standard deviations=0.35.Your research objective is to show that the population meanμexceeds2.5.Do the data provide sufficient evidence to conclude thatμ>2.5?(Use a 5% significance level.)H
₀is not rejected. There is insufficient evidence that to indicate that the mean is larger than 2.5.H
₀is rejected. There is insufficient evidence that to indicate that the mean is larger than 2.5.H
₀is rejected. There is sufficient evidence that to indicate that the mean is larger than 2.5.H
₀is not rejected. There is sufficient evidence that to indicate that the mean is larger than 2.5.You may need to use the appropriateappendix tableortechnologyto answer this question.

12.

[–/1 Points]DETAILSMENDSTAT15 9.2.022.MI.0/6 Submissions UsedMY NOTESASK YOUR TEACHERSuppose a scheduled airline flight must average at least62% occupancy in order to be profitable to the airline. Occupancy rates were recorded daily for a regularly scheduled flight on each of120 days,showing a mean occupancy per flight of60% and a standard deviation of10%.(a)Ifμis the mean occupancy per flight and if the company wishes to determine whether or not this scheduled flight is unprofitable, give the alternative and the null hypotheses for the test.H
₀:μ= 62 versusH
_a:μ≠62H
₀:μ≠62 versusH
_a:μ= 62H
₀:μH
_a:μ> 62H
₀:μ= 62 versusH
_a:μH
₀:μ= 62 versusH
_a:μ> 62(b)Does the alternative hypothesis in part (a) imply a one- or two-tailed test? Explain.Since only small values ofxwould tend to disprove the null hypothesis, this is a two-tailed test.Since only large values ofxwould tend to disprove the null hypothesis, this is a two-tailed test.Since only small values ofxwould tend to disprove the null hypothesis, this is a one-tailed test.Since only large values ofxwould tend to disprove the null hypothesis, this is a one-tailed test.Since small or large values ofxwould tend to disprove the null hypothesis, this is a two-tailed test.(c)Do the occupancy data for the 120 flights suggest that this scheduled flight is unprofitable? Test usingα= 0.05.(Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused region.)test statisticz=rejection regionz>zState your conclusion.H
₀is not rejected. There is insufficient evidence to indicate that the flight is unprofitable.H
₀is not rejected. There is sufficient evidence to indicate that the flight is unprofitable.H
₀is rejected. There is insufficient evidence to indicate that the flight is unprofitable.H
₀is rejected. There is sufficient evidence to indicate that the flight is unprofitable.You may need to use the appropriateappendix tableortechnologyto answer this question.

13.

[–/1 Points]DETAILSMENDSTAT15 9.2.023.MI.0/6 Submissions UsedMY NOTESASK YOUR TEACHERGround beef is packaged in small trays, intended to hold1 poundof meat. A random sample of30packagesin the small tray produced weight measurements with an average of1.02poundsand a standard deviation of0.14pounds.(a)If you were the quality control manager and wanted to make sure that the average amount of ground beef was indeed1 pound,what hypotheses would you test?H
₀:μH
_a:μ> 1H
₀:μ= 1 versusH
_a:μ> 1H
₀:μ= 1 versusH
_a:μH
₀:μ≠1 versusH
_a:μ= 1H
₀:μ= 1 versusH
_a:μ≠1(b)Find thep-value for the test and use it to perform the test in part (a). (Round your answer to four decimal places.)p-value =State your conclusion. (Useα= 0.05.)Since thep-value is greater than alpha,H
₀is rejected.Since thep-value is less than alpha,H
₀is not rejected.Since thep-value is less than alpha,H
₀is rejected.Since thep-value is greater than alpha,H
₀is not rejected.(c)How would you, as the quality control manager, report the results of your study to a consumer interest group?There is sufficient evidence to indicate that the average amount of ground beef is different from 1 pound.There is a 5% chance that any meat tray selected will weigh 1 pound.Since the sample average is greater than 1, all meat trays will weigh at least 1 pound.There is insufficient evidence to indicate that the average amount of ground beef is different from 1 pound.There is a 95% chance that any meat tray selected will weigh 1 pound.You may need to use the appropriateappendix tableortechnologyto answer this question.

14.

[–/1 Points]DETAILSMENDSTAT15 9.2.029.0/6 Submissions UsedMY NOTESASK YOUR TEACHERWhatisnormal, when it comes to people's body temperatures? A random sample of 130 human body temperatures had a mean of98.15°and a standard deviation of0.72°. Does the data indicate that the average body temperature for healthy humans deviates from 98.6°, the usual average temperature cited by physicians and others?(a)Test using thep-value approach withα= 0.05.State the null and alternative hypothesis.H
₀:μ= 98.6 versusH
_a:μ≠98.6H
₀:μ= 98.6 versusH
_a:μ> 98.6H
₀:μH
_a:μ> 98.6H
₀:μ≠98.6 versusH
_a:μ= 98.6H
₀:μ= 98.6 versusH
_a:μFind the test statistic and thep-value. (Round your test statistic to two decimal places and yourp-value to four decimal places.)z=p-value=State your conclusion.Thep-value is greater than alpha soH
₀is not rejected. There is insufficient evidence to indicate that the average body temperature for healthy humans deviates from 98.6°.Thep-value is less than alpha soH
₀is rejected. There is sufficient evidence to indicate that the average body temperature for healthy humans deviates from 98.6°.Thep-value is less than alpha soH
₀is rejected. There is insufficient evidence to indicate that the average body temperature for healthy humans deviates from 98.6°.Thep-value is greater than alpha soH
₀is not rejected. There is sufficient evidence to indicate that the average body temperature for healthy humans deviates from 98.6°.(b)Test using the critical value approach withα= 0.05.Find the rejection region. (Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused region.)z>zState your conclusion.The test statistic does not lie in the rejection region soH
₀is not rejected. There is insufficient evidence to indicate that the average body temperature for healthy humans deviates from 98.6°.The test statistic lies in the rejection region soH
₀is rejected. There is sufficient evidence to indicate that the average body temperature for healthy humans deviates from 98.6°.The test statistic lies in the rejection region soH
₀is rejected. There is insufficient evidence to indicate that the average body temperature for healthy humans deviates from 98.6°.The test statistic does not lie in the rejection region soH
₀is not rejected. There is sufficient evidence to indicate that the average body temperature for healthy humans deviates from 98.6°.(c)Compare the conclusions from parts (a) and (b). Are they the same?Yes, both methods produce the same conclusion.No, thep-value approach rejects the null hypothesis when the critical approach fails to reject the null hypothesis.No, the critical value approach rejects the null hypothesis when thep-value approach fails to reject the null hypothesis.(d)The 98.6°standard was derived by a German doctor in 1868, who claimed to have recorded 1 million temperatures in the course of his research.†What conclusions can you draw about his research in light of your conclusions in parts (a) and (b)?It appears as though the the doctor's equipment may not have been accurate.It appears as though the doctor was correct in his conclusion.You may need to use the appropriateappendix tableortechnologyto answer this question.

15.

[–/1 Points]DETAILSMENDSTAT15 9.2.030.MI.0/6 Submissions UsedMY NOTESASK YOUR TEACHERSome sports that involve a significant amount of running, jumping, or hopping put participants at risk for Achilles tendon injuries. A study looked at the diameter (in mm) of the injured tendons for patients who participated in these types of sports activities. Suppose that the Achilles tendon diameters in the general population have a mean of5.99millimeters(mm). When the diameters of the injured tendon were measured for a random sample of35patients,the average diameter was9.70mmwith a standard deviation of1.94mm. Is there sufficient evidence to indicate that the average diameter of the tendon for patients with Achilles tendon injuries is greater than5.99mm?Test at the5% levelof significance.State the null and alternative hypotheses.H₀:μ= 5.99 versusH_a:μ≠5.99H₀:μ= 5.99 versusH_a:μH₀:μ= 5.99 versusH_a:μ> 5.99H₀:μH_a:μ> 5.99H₀:μ≠5.99 versusH_a:μ= 5.99Find the test statistic and rejection region. (Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused region.)test statisticz=rejection regionz>zState your conclusion.H
₀is not rejected. There is sufficient evidence to indicate that the average diameter of the tendon for patients with AT is greater than 5.99 mm.H
₀is not rejected. There is insufficient evidence to indicate that the average diameter of the tendon for patients with AT is greater than 5.99 mm.H
₀is rejected. There is sufficient evidence to indicate that the average diameter of the tendon for patients with AT is greater than 5.99 mm.H
₀is rejected. There is insufficient evidence to indicate that the average diameter of the tendon for patients with AT is greater than 5.99 mm.You may need to use the appropriateappendix tableortechnologyto answer this question.

16.

[–/1 Points]DETAILSMENDSTAT15 9.3.001.0/6 Submissions UsedMY NOTESASK YOUR TEACHERIndependent random samples were selected from two quantitative populations, with sample sizes, means, and variances given below.

	Population
	1	2
Sample Size	38	45
Sample Mean	9.8	7.5
Sample Variance	10.83	16.49

State the null and alternative hypotheses used to test for a difference in the two population means.H
₀: (μ₁−μ₂)≠0 versusH
_a: (μ₁−μ₂) = 0H
₀: (μ₁−μ₂) = 0 versusH
_a: (μ₁−μ₂)
H
₀: (μ₁−μ₂) = 0 versusH
_a: (μ₁−μ₂) > 0H
₀: (μ₁−μ₂)
H
_a: (μ₁−μ₂) > 0H
₀: (μ₁−μ₂) = 0 versusH
_a: (μ₁−μ₂)≠0Calculate the necessary test statistic. (Round your answer to two decimal places.)z=Calculate the rejection region withα= 0.01.(Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused region.)z>zDraw the appropriate conclusion.H
₀is rejected. There is sufficient evidence to indicate a difference in mean.H
₀is rejected. There is insufficient evidence to indicate a difference in mean.H
₀is not rejected. There is sufficient evidence to indicate a difference in mean.H
₀is not rejected. There is insufficient evidence to indicate a difference in mean.You may need to use the appropriateappendix tableortechnologyto answer this question.

17.

[–/1 Points]DETAILSMENDSTAT15 9.3.004.NVA0/6 Submissions UsedMY NOTESASK YOUR TEACHERIndependent random samples were selected from two quantitative populations, with sample sizes, means, and variances given below.

	Population
	1	2
Sample Size	64	64
Sample Mean	4.0	5.2
Sample Variance	9.78	12.52

The value of the test statistics is−2.03.Calculate thep-value for the given data. (Round your answer to four decimal places.)p-value =Use thep-value to test for a significant difference in the population means at the 5% significance level.H
₀is rejected. There is sufficient evidence to indicate a difference in means.H
₀is not rejected. There is insufficient evidence to indicate a difference in means.H
₀is rejected. There is insufficient evidence to indicate a difference in means.H
₀is not rejected. There is sufficient evidence to indicate a difference in means.Is this result consistent with the one obtained using the critical value approach, withα= 0.05?YesNoYou may need to use the appropriateappendix tableortechnologyto answer this question.

18.

[–/1 Points]DETAILSMENDSTAT15 9.3.007.NVA0/6 Submissions UsedMY NOTESASK YOUR TEACHERIndependent random samples were selected from two quantitative populations, with sample data given below. Using thep-value approach for the data given below, is there sufficient evidence to show thatμ₁is larger thanμ₂at the 1% level of significance?

n1	=	n2= 50,x1=125.1,x2=123.6,
s1	=	5.5,s2=6.9

Use the value of the test statistics1.20to calculate thep-value for the test. (Round your answer to four decimal places.)p-value =State your conclusion.H
₀is not rejected. There is sufficient evidence to indicate that the mean forpopulation 1is larger than the mean forpopulation 2.H
₀is rejected. There is insufficient evidence to indicate that the mean forpopulation 1is larger than the mean forpopulation 2.H
₀is rejected. There is sufficient evidence to indicate that the mean forpopulation 1is larger than the mean forpopulation 2.H
₀is not rejected. There is insufficient evidence to indicate that the mean forpopulation 1is larger than the mean forpopulation 2.Is this result consistent with the one obtained using the critical value approach, withα= 0.01?YesNoYou may need to use the appropriateappendix tableortechnologyto answer this question.

19.

[–/1 Points]DETAILSMENDSTAT15 9.3.013.MI.0/6 Submissions UsedMY NOTESASK YOUR TEACHERAn experiment was planned to compare the mean time (in days) required to recover from a common cold for persons given a daily dose of 4 milligrams (mg) of vitamin C,μ₂, versus those who were not,μ₁. Suppose that32adults were randomly selected for each treatment category and that the mean recovery times and standard deviations for the two groups were as follows.

	No Vitamin Supplement	4 mg Vitamin C
Sample Size	32	32
Sample Mean	6.5	5.5
Sample Standard Deviation	2.8	1.1

(a)If you want to show that the use of vitamin C reduces the mean time to recover from a common cold, give the null and alternative hypotheses for the test.H
₀: (μ₁−μ₂)
H
_a: (μ₁−μ₂) > 0H
₀: (μ₁−μ₂) = 0 versusH
_a: (μ₁−μ₂) > 0H
₀: (μ₁−μ₂)≠0 versusH
_a: (μ₁−μ₂) = 0H
₀: (μ₁−μ₂) = 0 versusH
_a: (μ₁−μ₂)≠0H
₀: (μ₁−μ₂) = 0 versusH
_a: (μ₁−μ₂) Is this a one- or a two-tailed test?one-tailed testtwo-tailed test(b)Conduct the statistical test of the null hypothesis in part (a) and state your conclusion. Test usingα= 0.05.(Round your answer to two decimal places.)Find the test statistic.z=Find the rejection region. (Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused region.)z>zConclusion:H
₀is not rejected. There is sufficient evidence to indicate that Vitamin C reduces the mean recovery time.H
₀is not rejected. There is insufficient evidence to indicate that Vitamin C reduces the mean recovery time.H
₀is rejected. There is insufficient evidence to indicate that Vitamin C reduces the mean recovery time.H
₀is rejected. There is sufficient evidence to indicate that Vitamin C reduces the mean recovery time.You may need to use the appropriateappendix tableortechnologyto answer this question.

20.

[–/1 Points]DETAILSMENDSTAT15 9.3.015.0/6 Submissions UsedMY NOTESASK YOUR TEACHERAnalyses of drinking water samples for 100 homes in each of two different sections of a city gave the following information on lead levels (in parts per million).

	Section 1	Section 2
Sample Size	100	100
Mean	34.4	36.2
Standard Deviation	5.9	6.0

(a)Calculate the test statistic and itsp-value to test for a difference in the two population means. (UseSection 1−Section 2.Round your test statistic to two decimal places and yourp-value to four decimal places.)z=p-value=Use thep-value to evaluate the statistical significance of the results at the 5% level.H
₀is not rejected. There is insufficient evidence to indicate a difference in the mean lead levels for the two sections of the city.H
₀is not rejected. There is sufficient evidence to indicate a difference in the mean lead levels for the two sections of the city.H
₀is rejected. There is sufficient evidence to indicate a difference in the mean lead levels for the two sections of the city.H
₀is rejected. There is insufficient evidence to indicate a difference in the mean lead levels for the two sections of the city.(b)Calculate a 95% confidence interval to estimate the difference in the mean lead levels in parts per million for the two sections of the city. (UseSection 1−Section 2.Round your answers to two decimal places.)parts per million toparts per million(c)Suppose that the city environmental engineers will be concerned only if they detect a difference of more than5 partsper million in the two sections of the city. Based on your confidence interval in part (b), is the statistical significance in part (a) ofpractical significanceto the city engineers? Explain.Since all of the probable values ofμ₁−μ₂given by the interval are between−5 and 5, it is not likely that the difference will be more than 5 ppm, and hence the statistical significance of the difference is not of practical importance to the the engineers.Since all of the probable values ofμ₁−μ₂given by the interval are all less than−5, it is likely that the difference will be more than 5 ppm, and hence the statistical significance of the difference is of practical importance to the the engineers.Since all of the probable values ofμ₁−μ₂given by the interval are all greater than 5, it is likely that the difference will be more than 5 ppm, and hence the statistical significance of the difference is of practical importance to the the engineers.You may need to use the appropriateappendix tableortechnologyto answer this question.

21.

[–/1 Points]DETAILSMENDSTAT15 9.3.017.0/6 Submissions UsedMY NOTESASK YOUR TEACHERHow do states stack up against each other in SAT scores? To compare State 1 and State 2 scores, random samples of 100 students from each state were selected and their SAT scores recorded with the following results. (Useμ₁for State 1 andμ₂for State 2.)

State	Mean	Sample Size	Standard Deviation
State 1	1,123	100	192
State 2	1,049	100	167

(a)Use thecritical value approachto test for a significant difference in the average SAT scores for these two states at the 5% level of significance.State the null and alternative hypotheses.H
₀: (μ₁−μ₂)≠0 versusH
_a: (μ₁−μ₂) = 0H
₀: (μ₁−μ₂) = 0 versusH
_a: (μ₁−μ₂)
H
₀: (μ₁−μ₂) = 0 versusH
_a: (μ₁−μ₂) > 0H
₀: (μ₁−μ₂)
H
_a: (μ₁−μ₂) > 0H
₀: (μ₁−μ₂) = 0 versusH
_a: (μ₁−μ₂)≠0Find the test statistic. (Round your answer to two decimal places.)z=Find the rejection region. (Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused region.)z>zState your conclusion.H
₀is not rejected. There is sufficient evidence to indicate that there is a difference in the average SAT scores for the two states.H
₀is not rejected. There is insufficient evidence to indicate that there is a difference in the average SAT scores for the two states.H
₀is rejected. There is insufficient evidence to indicate that there is a difference in the average SAT scores for the two states.H
₀is rejected. There is sufficient evidence to indicate that there is a difference in the average SAT scores for the two states.(b)Use thep-value approachto test for a significant difference in the average SAT scores for these two states. (Useα= 0.05.)Find thep-value. (Round your answer to four decimal places.)p-value =If you were writing a research report, how would you report your results?The null hypothesis---Select---is notisrejected. There is---Select---sufficientinsufficientevidence to conclude that(μ₁−μ₂)?=≠>0.You may need to use the appropriateappendix tableortechnologyto answer this question.

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