The claim is that for 8 AM body temperatures of males, the mean is less than 98.6°F. The sample size is n = 36 and the test statistic is t= -3.675. Use technology to find the P-value. Based on the result, what is the final conclusion? Use a significance level of 0.01.State the null and alternative hypotheses.Ho: HVH₁: p V(Type integers or decimals. Do not round. Do not include the degree symbol in your answer.)The test statistic is.(Round to two decimal places as needed.)The P-value is.(Round to three decimal places as needed.)Based on the P-value, there ▼sufficient evidence at a significance level of 0.01 toG▼the claim that for 8 AM body temperatures of males, the mean is less than 98.6°F.

The claim is μ<98.6. Given that n=36, t=-3.675, α=0.01.

Answered: The claim is that for 8 AM body…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Use technology to find the P-value for the hypothesis test described below. The claim is that for 12 AM body temperatures, the mean is µ> 98.6°F. The sample size isn=9 and the test statistic is t=2.568. P-value = (Round to three decimal places as needed.)
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Consumer Reports did a survey of 1100 adults and found that 565 were unhappy that their food did not look like the picture in the menu. Use a 0.05 significance level to test the claim that the majority of adults are annoyed by food that does not look like the picture in the menu. Identify the null and alternate hypotheses. TTTArial 3 (12pt) v T- = - E·V
Determine P-Value.
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The data in the accompanying table give the weights (in g) of randomly selected quarters that were minted after 1964. The quarters are supposed to have a median weight of 5.670 g. Use the sign test and a 0.01 significance level to test the claim that the median is equal to 5.670 g. Do quarters appear to be minted according to specifications? Click the icon to view the data. What are the null and alternative hypotheses? O A. Ho: Median weight = 5.670 g H₁: Median weight 5.670 g More Info OB. Ho: Median weight ≤ 5.670 g H₁: Median weight = 5.670 g O D. Ho: Median weight = 5.670 g H₁: Median weight # 5.670 g 0 Post-1964 Quarters 5.5863 5.5764 5.6582 5.5649 5.5844 5.7402 5.7367 5.6057 5.5995 5.6165 5.6448 5.5974 5.6609 5.6363 5.6352 5.5986 5.5797 5.5501 5.6238 5.6462 5.6423 5.7021 5.5481 5.7078 5.6055 5.5324 5.7417 5.6776 5.5924 5.5691 5.5666 5.7287 5.6312 5.6017 5.6626 5.6436 5.6634 5.6132 5.5441 5.6363 -
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Use the following R output to answer the question. Use the following R output to answer the question. >chisq.test(data) Pearsons Chi-square test data:data X-squared=4.7194, df=2, p-value=0.09445 If the significance level is 5%, what conclusion can you make? Group of answer choices A. fail to reject the alternative hypothesis B. cannot determine C. fail to reject the null hypothesis D. reject the null hypothesis and accept the alternative hypothesis
Westjet claims that it's avearage time to fly to Toronto from Thunder Bay is 121 minutes. A random sample of 29 flights from Thunder Bay to Toronto were observed and yeilded a mean of 140 minutes. Assuming that o = 3.1 do we have enough evidence to show that the mean flight is more than 121 minutes? Use a = 0.1 and use our tables. Select the correct null and alternative hypotheses: OA. Ho: X= 140,HA: X 121 OF. Ho X = 140,H₁ : X > 140 OG. None of the above Select the type of distribution to be used in this question: OA. Z-distribution B. t-distribution OC. Chi-Square distribution OD. None of the above The rejection region for this test is: OA. (-∞, -1.28) OB. (1.96,∞) OC. (1.645, ∞) OD. (1.28, ∞) OE. (-∞, -1.96) OF. (-∞, -1.645) OG. (-∞, -1.96) U (1.96,∞) OH. (-∞, -1.28) U (1.28,00) OI. (-∞, -1.645) U (1.645, ∞) OJ. None of the above Test statistic = The conclusion for this test is: OA. Fail to Reject Ho OB. Reject Ho OC. Fail to Reject HA OD. Reject HA OE. None of the above
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