Part II: Hypotheses: Calories of the same students will be significantly different based on theterm typeQ1. What test did you use to assess hypothesis?Q2. What are the 2 assumptions that should be tested on SPSS? Can we make theseassumptions based on the given data? No need to provide figuresQ3. Is the above hypothesis true? Explain by providing the P valueQ4. Is there a need for post-hoc tests here? Why or why not. If yes, copy, paste and interpret theresult from the post-hoc test to this Word doc. If "No", do not copy anything at allFall Term245344451251145352547552452251446542142143Spring Term141145239141546337548445148337444141530435Summer Term233133320528435223348430297439386438272397

Answered: Part II: Hypotheses: Calories of the…

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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You wish to test the following claim (H) at a significance level of a = Ho: P1 = P2 Ha: P₁ P2 You obtain a sample from the first population with 122 successes and 438 failures. You obtain a sample from the second population with 63 successes and 304 failures. test statistic = p-value = [three decimal accuracy] 0.001. [four decimal accuracy]
Answer the following statistic questions on house prices, use the information below as reference.
also option We should not reject the null hypothesis because t0>tα. So, at the 10% significance level, the data do not provide sufficient evidence to conclude that the average length of time between dentist visits is less than 193days.
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Using the results obtained for a study on website development below, how would we interpret the results of this hypothesis test? The claim is the percent of customers who abandon a web page due to long load times is different from 34%. The significance level is 0.10. The test statistic is 2.50. Select the correct answer below: Reject H0. There is strong evidence to conclude that the percent of customers who abandon a web page due to long load times is greater than 34%. Reject H0. There is NOT strong evidence to conclude that the percentage of customers who abandon a web page due to long load times is greater than 34%. Fail to reject H0. There is strong evidence to conclude that the percent of customers who abandon a web page due to long load times is greater than 34%. Fail to reject H0. There is NOT strong evidence to conclude that the percentage of customers who abandon a web page due to long load times is greater than 34%.
5.32 Fuel efficiency of manual and automatic cars, Part I: Do these data provide strong evidence of a difference between the average fuel efficiency of cars with manual and automatic transmissions in terms of their average city mileage? Assume that conditions for inference are satisfied. City MPG, Automatic City MPG, Manual Mean 16.12 19.85 SD 3.58 4.51 n 26 26 The test statistic is:The p-value is:
Since an instant replay system for tennis was introduced at a major tournament, men challenged 1413 referee calls, with the result that 426 of the calls were overturned. Women challenged 742 referee calls, and 228 of the calls were overturned. Use a 0.05 significance level to test the claim that men and women have equal success in challenging calls. Identify the test statistic. z= ? (Round to two decimal places as needed.) Part 3 Identify the P-value. P-value= ? (Round to three decimal places as needed.) Part 4 What is the conclusion based on the hypothesis test? The P-value is ▼ greater than or less than the significance level of α=0.05, so ▼ fail to reject or reject the null hypothesis. There ▼ is not sufficient or is sufficient evidence to warrant rejection of the claim that women and men have equal success in challenging calls. Part 5 b. Test the claim by constructing an appropriate confidence interval. The 95% confidence…
Dlana- Outlook CS Diana CougarWeb S My Orders | SHEIN Question 44 Test the claim that the proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.05 significance level. The null and alternative hypothesis would be: Ho:HM HF H1: µM > HF Hị:PM PF H1:pPM PF H1: µM # HF Ho: HM 2 HF Ho: PM < PF Ho: PM = PF Ho:HM = HF The test is: right-tailed two-tailed left-tailed Based on a sample of 80 men, 45% owned cats Based on a sample of 60 women, 65% owned cats test statistic = [three decimal accuracy] p-value = [three decimal accuracy] Based on this we: O Reject the null hypothesis O Fail to reject the null hypothesis 10 MacBook Air 80 000 000 esc F2 F3 F4 F5 F6 F1
Looking at youth depression scores from a general sample of random children in the US, we want to determine if our sample is significantly different than the national average of scores of children with depression in the US. The national average of scores on a measure of depression is 5 points. We have no reason to assume it will be higher or lower than the US national average. The sample: 11 9 12 3 3 4 13 1. What is the null and alternative hypotheses? Do not specify any directionality. 2. What did you discover (using 95% Cl)? Provide an interpretation.
Tests for adverse reactions to a new drug yielded the results given in the table. At the 0.05 significance level, researcher wants to test if the data provide sufficient evidence to conclude that an association exists between the treatment (drug or placebo) and the reaction (whether or not headaches were experienced)? Question 1 Write the null and the alternative hypothesis. Question 2 Calculate the test statistic. Question 3 Calculate the critical value.. Question 4 What is the decision of the hypothesis testing. Drug Placebo Headaches 11 7 No headaches 73 91
Three students, an athlete, a fraternity member, and an honors student, record the number of hours they slept each night for 20 nights. O JMP Applet imp ? Oneway Analysis of Sleep Hours By Student Oneway Analysis of Sleep Hours By Student 10 Oneway Anova 14 Summary of Fit 12 Rsquare 0.024506 10- Adj Rsquare Root Mean Square Error Mean of Response Observations (or Sum Wgts) -0.00072 1.99517 7.7 60 Analysis of Variance Sum of Mean F Prob > Source DF Squares Square Ratio F 2 Athiete Frat Honors Student 2 5.70000 2.85000 0.7180 0.4931 Student Error 57 226.90000 3.98070 C. Total 59 232.60000 Oneway Anova Means for Oneway Anova Std Lower Upper 95% Summary of Fit Level Number Mean Error 95% Athlete 20 8.10000 0.44813 7.2086 8.9934 Rsquare 0.024506 Frat 20 7.65000 0.44813 6.7588 8.5434 Adj Rsquare Root Mean Square Error Mean of Response Observations (or Sum Wgts) -0.00972 Honors 20 7.35000 0.44813 6.4586 8.2434 1.99517 Std Error uses a pooled estimate of error variance 7.7 60 Analysis of…
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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