If, in a one-tail ypothesis test where H, is only rejected in the upper tail, the p value = 0.0144 and ZgTAT+219, what is the statistical decision if the nul hypothesis is tested at the0.02 level of significance?What is the statistical decsion?HeSince the p-value is(Type an integer or agreater than or equal toless than

GivenP-value=0.0144Zstat=+2.19Level of significance(α)=0.02

Answered: If, in a one tail typothesis test where…

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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Assume that matched pairs of data from a poll of Internet users who were asked if they make travel plans through the Internet result in 304 positive signs, 278 negative signs, and 35 ties when the value of the secon variable is subtracted from the corresponding value of the first variable. Use the sign test with a 0.05 significance level to test the null hypothesis of no difference. Let ₁ denote the median of the first variable and 12 denote the median of the second variable. What are the null and alternative hypotheses? O A. Ho: 1₁2 112 H₁:1₁ <1₂ OC. Ho: 11₁ 112 H₁: 1₁ <12 O B. Ho: 11 #11₂ H₁:1₁ =1₂ O D. Ho: 111 112 H₁:1₁ #1₂ Find the test statistic. Test statistic = Find the P-value. P-value= (Round to four decimal places as needed.) Does the null hypothesis of no difference hold? (Round to two decimal places as needed.) O A. Since the P-value is less than the significance level, reject the null hypothesis of no difference. O B. Since the P-value is greater than the significance…
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Whatt is the long hand solution for You select a sample of n =36 participants and aask them to walk every day for at least 30 minutes. At the end of a month you give them the well-being scale and find M = 72. If you use a two-tailed test with alpha level equal to 0.05 to determine the impactof taking a daily walk on well-being, what is the probability that you have made a Type I error? How can you reduce the probability of making a Type I error?
6.6.16-T Question Help The probability of flu symptoms for a person not receiving any treatment is 0.029. In a clinical trial of a common drug used to lower cholesterol, 33 of 1011 people treated experienced flu symptoms. Assuming the drug has no effect on the likelihood of flu symptoms, estimate the probability that at least 33 people experience flu symptoms. What do these results suggest about flu symptoms as an adverse reaction to the drug? (a) P(X2 33) = (Round to four decimal places as needed.) Enter your answer in the answer box and then click Check Answer. 1 part remaining Check Answer Clear All 41,109
Several students were tested for reaction times (in thousandths of a second) using their right and left hands. (each value is the elapsed time between the release of a strip of paper and the instant that it is caught by the subject.) REsults from five of the students are includedin the graph. Use a significance level to test the claim that there is no difference between the reaction times on the right and left hands.What are the hypothesis for this test?What is the test satistic? Identify the Critical Value(s)What is the Conclusion?
COMPUTE the standard error of the mean?
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
Are blonde female college students more likely to have boyfriends than brunette female college students? 219 of the 350 blondes surveyed had boyfriends and 220 of the 400 brunettes surveyed had boyfriends. What can be concluded at the 0.05 level of significance? H0: PBlonde = PBrunette Ha: PBlonde [ Select ] ["Not Equal To", ">", "<"] PBrunette Test statistic: [ Select ] ["Z", "T"] p-Value = [ Select ] ["0.04", "0.16", "0.08", "0.02"] [ Select ] ["Reject Ho", "Fail to Reject Ho"] Conclusion: There is [ Select ] ["insufficient", "statistically significant"] evidence to make the conclusion that blonde female college students are more likely to have boyfriends than…
A local chess club claims that the length of time to play a game has a mean of 23 minutes or more. Write sentences describing type I and type II errors for a hypothesis test of this claim. A type I error will occur if the actual mean of the length of time to play a game is ▼ less than equal to less than or equal to greater than not equal to greater than or equal to 23 minutes, but you ▼ fail to reject reject the null hypothesis, ▼ Upper H 0 : mu less than 23H0: μ<23 Upper H 0 : mu equals 23H0: μ=23 Upper H 0 : mu not equals 23H0: μ≠23 Upper H 0 : mu less than or equals 23H0: μ≤23 Upper H 0 : mu greater than or equals 23H0: μ≥23 Upper H 0 : mu greater than 23H0: μ>23 . A type II error will occur if the actual mean of the length of time to play a game is ▼ not equal to greater than or equal to equal to greater than less than or equal to less than 23 minutes, but you ▼ fail to reject reject the null hypothesis, ▼…
This is the result of Chi Square test between 2 variables: gender and length of service. If the Null Hypothesis is stated as "There is no significant association between gender and length of service", then table above will clearly allow us to conclude that "there is no significant association between gender and length of service". A. True B. False
You are conducting a multinomial Goodness of Fit hypothesis test for the claim that all 5 categories are equally likely to be selected. Complete the table. Report all answers correct to three decimal places. H0:pA=pB=pC=pD=pEH0:pA=pB=pC=pD=pEHa:Ha:At least one probability is not equal to the others (claim) Expected frequency 15.6 1.) What is the chi-square test-statistic for this data?χ2=
The correlation coefficient r is a sample statistic. What does it tell us about the value of the population correlation coefficient ? (Greek letter rho)? You do not know how to build the formal structure of hypothesis tests of ? yet. However, there is a quick way to determine if the sample evidence based on ? is strong enough to conclude that there is some population correlation between the variables. In other words, we can use the value of r to determine if ? ≠ 0. We do this by comparing the value |r| to an entry in the correlation table. The value of ? in the table gives us the probability of concluding that ? ≠ 0 when, in fact, ? = 0 and there is no population correlation. We have two choices for ?: ? = 0.05 or ? = 0.01. (a) Look at the data below regarding the variables x = age of a Shetland pony and y = weight of that pony. Is the value of |r| large enough to conclude that weight and age of Shetland ponies are correlated? Use ? = 0.05. (Round your answer for r to four decimal…

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