Consider a drug that is used to help prevent blood clots in certain patients. In clinical trials, among 6034 patients treated with this drug, 165 developed the adverse reaction of nausea. Use a 0.10significance level to test the claim that 3% of users develop nausea. Does nausea appear to be a problematic adverse reaction?Identify the null and alternative hypotheses for this test. Choose the correct answer below.OA. Ho: p=0.03H₁: p=0.03OB. Ho: p=0.03H₁: p<0.03OC. Ho: p=0.03H₁: p>0.03OD. Ho: p *0.03H₁: p=0.03Identify the test statistic for this hypothesis test.The test statistic for this hypothesis test is(Round to two decimal places as needed.)Identify the P-value for this hypothesis test.The P-value for this hypothesis test is.(Round to three decimal places as needed.)Identify the conclusion for this hypothesis test....O A. Reject Ho. There is sufficient evidence to warrant rejection of the claim that 3% of users develop nausea.OB. Fail to reject Ho. There is not sufficient evidence to warrant rejection of the claim that 3% of users develop nausea.

From the information, given thatThe claim is to test whether the proportion of users develop nausea…

Answered: Consider a drug that is used to help…

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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Test the claim that the mean GPA of night students is significantly different than 2.4 at the 0.2 significance level.The null and alternative hypotheses would be: H0:p=0.6H0:p=0.6H1:p≠0.6H1:p≠0.6 H0:μ=2.4H0:μ=2.4H1:μ≠2.4H1:μ≠2.4 H0:μ=2.4H0:μ=2.4H1:μ>2.4H1:μ>2.4 H0:p=0.6H0:p=0.6H1:p<0.6H1:p<0.6 H0:μ=2.4H0:μ=2.4H1:μ<2.4H1:μ<2.4 H0:p=0.6H0:p=0.6H1:p>0.6H1:p>0.6 The test is: right-tailed left-tailed two-tailed Based on a sample of 31 people, the sample mean GPA was 2.36 with a standard deviation of 0.16The p-value is: (to 3 decimals)The positive significance level is: (to 2 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesis
Identify the type I error and the type II error that corresponds to the given hypothesis. The proportion of adults who use the internet is greater than 0.27. Which of the following is a type I error? O A. Reject the claim that the proportion of adults who use the internet is 0.27 when the proportion is actually 0.27. O B. Fail to reject the claim that the proportion of adults who use the internet is 0.27 when the proportion is actually different from 0.27. O C. Reject the claim that the proportion of adults who use the internet is 0.27 when the proportion is actually different from 0.27. O D. Fail to reject the claim that the proportion of adults who use the internet is 0.27 when the proportion is actually 0.27. Which of the following is a type II error? O A. Fail to reject the claim that the proportion of adults who use the internet is 0.27 when the proportion is actually 0.27. O B. Reject the claim that the proportion of adults who use the internet is 0.27 when the proportion is…
A statistical program is recommended. An article reported data on potential drop measurements for one sample of connectors wired with alloy aluminum and another sample wired with EC aluminum. Does the accompanying computer output suggest that the true average potential drop for alloy connections (type 1) is higher than that for EC connections (as stated in the article)? Carry out the appropriate test using a significance level of 0.01. [Note: The computer output reports the P-value for a two-tailed test.] (Use μ₁ for alloy connections and μ2 for EC connections.) Type 1 2 O Ho: M₁ Ha: M₁ Variances State the relevant hypotheses. Ho: M₁ M₂ = 0 Ha: M1 - M2+ 0 Ho: M₁ M₂ = 0 На: M1 - M2 0 O Ho: M₁ M₂ > 0 Ha: M1 - M2=0 Mean 19.49000000 18.90000000 DF 39.5 40.0 Std Dev 0.55012821 0.48998389 Prob>|T| 0.0007 0.0007 Std Error 0.12004782 0.10692325 Find the test statistic and P-value using the output. (Enter your answers to four decimal places.) t = P-value =
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State the Result: A hypothesis test was conducted at the alpha = 0.01 level of significance. The test resulted in a p-value of 0.044.
A survey across generations of workers gathered data on engagement at work. The results for a sample of 1,000 workers are in the accompanying contingency table. At the 0.10 level of significance, is there evidence of a significant relationship between generation and level of engagement in the workplace? E Click the icon to view the contingency table. What are the null and alternative hypotheses to test? H,: No relationship exists between generation and level of engagement. H,: A relationship exists between generation and level of engagement. Generation and level of engagement Calculate the test statistic. XŠTAT =O GENERATION LEVEL OF ENGAGEMENT Millennials Gen Xers Baby Boomers Traditionalists Total Engaged Not Engaged Actively Disengaged Total (Round to two decimal places as needed.) 104 111 97 16 328 188 165 127 18 498 59 60 50 5 174 351 336 274 39 1,000 Print Done
a study of store checkout scanners, 1234 items were checked and 23 checked items were overcharges. Use a 0.05 significance level to test the claim that with scanners, 1% of sales are overchanrges. (Before scanners were used, the overcharge rate was estimated to be about 1%). a. Define the parameter A. mu = The proportion of all sales that are undercharges B. p = The proportion of all sales that are incorrect C. p = The proportion of all sales that are overcharges D. mu = The mean number of sales that are overcharges b. State the null and alternative hypotheses A. Upper H 0 : mu not equals 0.01 Upper H 1 : mu equals 0.01 B. Upper H 0 : p greater than 0.01 Upper H 1 : p equals 0.01 C. Upper H 0 : p equals 0.01 Upper H 1 : p not equals 0.01 D. Upper H 0 : p equals 23 Upper H 1 : p less than 23 c. Calculate the sample proportion ModifyingAbove p…
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 0.34. You use a significance level of α=0.005α=0.005. H0:p=0.34H0:p=0.34 H1:p≠0.34H1:p≠0.34You obtain a sample of size n=551n=551 in which there are 164 successes.What is the test statistic for this sample? (Report answer accurate to three decimal places.)test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.)p-value =
Next Question Assume a significance level of a = 0.05 and use the given information to cd d (b) below. Original claim: More than 43% of adults would erase all of their personal information online if they could. The hypothesis test results in a P-value of 0.0443. ..... a. State a conclusion about the null hypothesis. (Reject Ho or fail to reject Ho.) Choose the correct answer below. A. Fail to reject Ho because the P-value is greater than x. B. Reject H, because the P-value is less than or equal to a. C. Fail to reject H, because the P-value is less than or equal to a. O D. Reject Ho because the P-value is greater than a. b. Without using technical terms, state a final conclusion that addresses the original claim. Which of the following is the correct conclusion? A. There is sufficient evidence to support the claim that the percentage of adults that would erase all of their personal information online if they could is more than 43%. B. There is not sufficient evidence to support the…
A health manager wants to assess the relationship between number of doctor visits and patient feedbacks.Patient feedbacks are categorized between 1= very bad and 5= very goodin a scale. The SPSS analysis is given below. Write down the hypothesis for this case and specify the test used and comment on the results. Patient Feedback Visit to Doctor 5.00 2.00 4.00 1.00 4.00 2.00 5.00 3.00 4.00 3.00 3.00 4.00 3.00 4.00 4.00 4.00 2.00 5.00 2.00 5.00 3.00 5.00 1.00 4.00 One-Sample Kolmogorov-Smirnov Test Kisisel Saglik Degerlendirmesi Gecen Yil Icindeki Hekim Ziyaret Sayisi N 12 12 Normal Parametersa,b Mean 3.3333 3.5000 Std. Deviation 1.23091 1.31426 Most Extreme Differences Absolute .206 .232 Positive .127 .127 Negative -.206 -.232 Kolmogorov-Smirnov Z .713 .802 Asymp. Sig. (2-tailed) .689 .541 a. Test…
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