KTrials in an experiment with a polygraph include 96 results that include 23 cases of wrong results and 73 cases of correct results. Use a 0.05 significance level to test the claim that such polygraphresults are correct less than 80% of the time. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses theoriginal claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution.cxA@Let p be the population proportion of correct polygraph results. Identify the null and alternative hypotheses. Choose the correct answer below.OA. Ho: p=0.20H₁: p<0.202View an example Get more help.OE. Ho: p=0.20H₁: p*0.20*WSc. Ho: p=0.80H₁: p<0.80The test statistic is z= -0.97. (Round to two decimal places as needed.)The P-value is. (Round to four decimal places as needed.)Monroe's Temp....docxRemovedXmmandH#380EDC$4AQRFSAMPLE Persuas....pdf%5VTG6ABMacBook AirFOYHD 5373255 (1) (1).docx&7UNOB. Ho: p=0.20H₁: p>0.20JD. H₂ p=0.80H₁: p*0.80OF H₂: p=0.80H₁: p>0.80*8ADII1MD 5373255 (1).docx9KO1двO4F10P7Lcommand>Clear all.:;d{+[option?Check answer1111Show AllF}1Xdelete

Given: Sample size (n) = 96 From which 73 cases of correct results. Significance level(α) = 0.05

Answered: Trials in an experiment with a…

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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Do men score the same on average compared to women on their statistics finals? Final exam scores of eleven randomly selected male statistics students and thirteen randomly selected female statistics students are shown below. Male: 70 81 94 62 73 84 87 57 92 76 75 Female: 86 48 66 80 46 74 66 67 66 76 82 46 67 Assume both follow a Normal distribution. What can be concluded at the the a = 0.01 level of significance level of significance? For this study, we should use Select an answer a. The null and alternative hypotheses would be: Ho: Select an answer v Select an answer v Select an answer v (please enter a decimal) H1: Select an answer v Select an answer v Select an answer v (Please enter a decimal) b. The test statistic ? v (please show your answer to 3 decimal places.) c. The p-value = (Please show your answer to 4 decimal places.) d. The p-value is ? v a e. Based on this, we should Select an answer v the null hypothesis. f. Thus, the final conclusion is that ... O The results are…
Manufacturer of a CNC machine claims that the average warm-up time for their machine is 10 minutes. We would like to test this claim at a 0.10 significance level and therefore run 10 trials that result in the folowing warnm-up durations (in minutes). Warm-up durations are assumed to follow a Normal distribution. 9.6, 9.9, 10.5. 10.4, 9.7, 9.8, 9.6, 10.3, 10.1, 9.6 Click here to view page 1 of the table of critical values of the t-distribution. Click here to view page 2 of the table of critical values of the t-distribution. (a) Choose the correct Hg and H, below. O A. H,: H> 10 against H, : u= 10 O B. H,: p= 10 against H, : u 1o O D. H: µ# 10 against H,: u= 10 O E. H,: µ OB. t (d) Choose the correct conclusion below. O A. Do not reject the null hypothesis because the average warm-up time is not significantly different from 10 minutes. O B. Reject the null hypothesis because the average wam-up time is not significantly different from 10 minutes. OC. Reject the null hypothesis because the…
The amount of water consumed each week by Bronx residences is normally distributed. The population mean of water consumption is 120.3 gallons with a standard deviation of 10.0 gallons. Test the claim at the 0.10 significance level that the average amount of water consumed is not 125 gallons with a sample size of 100 residences. Do you retain or reject the null hypothesis?
A study was performed concerning risk factors for carotid-artery stenosis (arterial narrowing) among 264 men. The data reported is: Sample size Mean Standard deviationGroup 1: No stenosis 156 5.3 1.4Group 2: Stenosis 108 5.1 0.8 The test stastistic F is assumed to be 3.06 to check the equal variance. Question: Suppose researchers want to test if there is a significant difference in mean blood-glucose level between men with and without stenosis considering the critical F test is 1.35 with a equal variance (pooled), what is the test statistic for this test ?
Scores on a standarized exam are known to follow a normal distribution with a standard deviation of 5. A researcher finds a random sample of 25 exam scores has a mean score of 72. Use this to find a 88% confidence interval for the mean exam score of all students taking the exam.
Small-business telephone users were surveyed 6 months after access to carriers other than Carrier A became available for wide-area telephone service. Of a random sample of 240 Carrier A users, 135 said they were attempting to learn more about their options, as did 227 of an independent random sample of 260 users of alternate carriers. Test, at the 1% significance level against a two-sided alternative, the null hypothesis that the two population proportions are the same. Click the link to view a standard normal distribution table. be the Let Px be the proportion of Carrier A users who said they were attempting to learn more about their options and Py proportion of users of alternate carriers who said they were attempting to learn more about their options. Determine the null and alternative hypotheses. Choose the correct answer below. O A. Ho: Px-Py=0 H₁: Px-Py 20 O D. Ho: Px-P₁ #0 H₁: Px - Py = 0 B. Ho: Px-Py = 0 H₁: Px-Py 0
In a random sample of males, it was found that 21 write with their left hands and 210 do not. In a random sample of females, it was found that 56 write with their left hands and 440 do not. Use a 0.05 significance level to test the claim that the rate of left-handedness among males is less than that among females. Identify the test statistic. and p value
A researcher wishes to test whether the mean of a normally distributed population is greater than 1.06 at the 5% significance level. Based upon 22 independent observations, she observes a sample mean of 1.77 and a sample standard deviation of 0.84. Determine the degrees of freedom of the null distribution of Student's t test.
3 - A water manufacturer claims that its product is 330 ML. The amount of water has a normal distribution. Quality manager takes 25 products every hour and calculate the amount of the water. The mean amount of water at 10:00 yields the average amount of 315 ML and variance of 9 ML. Test the claim that amount of the water in the bottles is at least 330 ML. Use significance level as 0.05. What is the hypothesis? What is the test statistics? If both of the answers are correct you got the points, otherwise you lose all points. (use two decimal) a ) 25.00 b) Но : и — 330 Н) : д + 330 c) Но : и 330 d) -8.33 e) O Ho : µ 330 f) 8.33 g) Η, : μ>330 Η : μ 330 O O O 0 0
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In an exam imposed by a corporation on a very large number of people, we consider that the score obtained isa random variable obeying a normal law. Note mean is 65 and variance is 152.The corporationdecides that only 25% of participants will be accredited by this exam. What grade do you need to get to beaccredited?
Workers' Distractions. A recent study showed that the modern working person experiences an average of 2.1 hours per day of distractions (phone calls, e- mails, impromptu visits, etc.). A random sample of 16 workers for a large corporation found that these workers were distracted an average of 1.8 hours per day and the standard deviation was 20 minutes. Estimate the true mean population distraction time with 90% confidence, and compare your answer to the results of the study. (1.654,1.946). Mean of the study is outside or "different" from the result of our study.

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