In a program designed to help patients stop smoking, 198 patients were given sustained care, and 82.0% of them were no longer smoking after one month. Use a 0.05significance level to test the claim that 78% of patients stop smoking when given sustained care. Use the P-value method. Use the normal distribution as anapproximation to the binomial distribution.The Hypotheses test: Choose one of these( answer A or B or C)А. Но: р - 0.78В. Но: р - О.78С. Но: р - О.78H1:p > 0.78H1:P+ 0.78H1:p < 0.78Test statistic:Z =P value =Conclusion: since p value( answer: greater or less than ) significance level a = 0.05|( answer: Fail to reject or Reject) the null hypotheses.

Introduction: Consider that p is the true proportion of patients who stop smoking when given…

Answered: In a program designed to help patients…

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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Before the furniture store began its ad campaign, it averaged 123 customers per day. The manager is investigating if the average is larger since the ad came out. The data for the 11 randomly selected days since the ad campaign began is shown below: 110, 154, 108, 112, 151, 146, 144, 130, 138, 129, 142 Assuming that the distribution is normal, what can be concluded at the a = 0.05 level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho: vSelect an answer H1: ?V Select an answer V c. The test statistic ? (please show your answer to 3 decimal places.) d. The p-value = e. The p-value is ?V a f. Based on this, we should Select an answer g. Thus, the final conclusion is that ... (Please show your answer to 4 decimal places.) the null hypothesis. O The data suggest the populaton mean is significantly more than 123 at a = 0.05, so there is sufficient evidence to conclude that the population mean number of customers since the…
a. 14.0, 16.3, 18.7, 19.9, 21.3 b. 16.1, 13.4, 13.6, 14.9, 14.1, 10.5 Construct the 95% confidence interval for the difference in u of the two given samples.
Part: 2/5 Part 3 of 5 (c) Compute the P-value. Round the answer to at least four decimal places. P-value= X Ś
Zachary works as a quality control engineer at Ford Motor Company. One of his job responsibilities is estimating the proportion of defective engines coming off a production line. The production line setting allows for a defective rate of 5%. In a periodical quality checkup, Zachary selected 100 engines that came off the production line. What is the chance Zachary will find more than 10 defective engines? The corresponding Z score is _____
A supplier of digital memory cards claims that no more than 1% of the cards are defective. In a random sample of 600 memory cards, it is found that 3% are defective, but the supplier claims that this is only a sample fluctuation. How many populations? 01 What is the parameter? O tandard Deviation Variance Difference between Means Proportion Mean
Which of these scenarios means that a pretest-posttest design with a non-equivalent control group is not suitable? Select all that apply. O a. All of the available participants will be receiving the intervention at the same time. O b. Authors want to examine for any trends in the data over time. O c. The participants were placed into the experimental and control groups by natural circumstances rather than random assignment. O d. Authors want to compare averages between two groups before and after the intervention using an ANOVA.
In a one-way analysis of variance with three treatments, each with five measurements, in which a completely randomized design is used, what is the degrees of freedom for treatments? a. 5 b. 2 c. 4 d. 8
Listed below are the lead concentrations (in ug/g) measured in different Ayurveda medicines. Ayurveda is a traditional medical system commonly used in India. The lead concentrations listed here are from medicines manufactured in the United States. Assume that a simple random sample has been selected. Use a 0.01 significance level to test the claim that the mean lead concentration for all such medicines is less than 14.0 µg /g. 2.96 6.45 5.99 5.51 20.53 7.45 11.97 20.46 11.52 17.54 D Identify the null and alternative hypotheses. Ho: H1: (Type integers or decimals. Do not round.) Identify the test statistic. (Round to two decimal places as needed.) Identify the P-value. (Round to three decimal places as needed.) State the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. V the null hypothesis. There sufficient evidence at the 0.01 significance level to V the claim that the mean lead concentration for all Ayurveda medicines…
The coach of a very popular men’s basketball team claims that the average distance the fans travel to the campus to watch a game is 35 miles. The team members feel otherwise. A sample of 16 fans who travel to games was randomly selected and yielded a mean of M= 36 miles and s= 5 miles. Test the coach’s claim at the 5% (.05) level of significance. one-tailed or two-tailed test: State the hypotheses: df= tα or t value for the critical region = sM = t (test statistic)= Decision:
For a two-tailed test with a sample size of 17 and a 0.20 level of significance, the t value is O a. 1.337 O b. 0.865 O c. 1.230 d. 1.734 O
The accompanying table lists pulse rates. Use a 0.05 significance level and apply the methods of two-way analysis of variance. What is the conclusion? Click on the icon to view the data table. C State the null and alternative hypotheses in the test for the effect of an interaction between row and column factors. Ho: There is no interaction between gender and age. H₁: There is an interaction between gender and age. What is the value of the test statistic for this test? F = 3.32 (Round to two decimal places as needed.) What is the corresponding P-value of the test statistic, F, for this P-value= (Round to three decimal places as needed.) Pulse Rates for Gender and Age Over 30 Years of Age Under 30 Years of Age Female 78 103 78 63 61 98 81 98 91 95 77 75 73 65 71 78 61 71 74 55 Female Male 60 81 56 69 68 74 75 68 63 56 46 70 61 65 91 80 59 58 64 59 Male D X
Which outcome is expected if the null hypothesis is true for an analysis of variance? a. SSbetween should be about the same size as SStotal. b. SSbetween should be about the same size as SSwithin. c. MSbetween should be about the same size as MStotal. d. MSbetween should be about the same size as MSwithin.

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