10 for both samplesQUESTION 7A researcher reports t(12) = 2.86, p=.04 for a repeated-measures research study. From this we know:Oa. The researcher would have failed to reject the null if their alpha = .05, and there were 12 participants in the sample.Ob. The researcher would have rejected the null, their alpha = 2.86, and there were 13 participants in the sample.OC. The researcher would have rejected the null if their alpha = .05, and there were 13 participants in the sample.d. The researcher would have failed to reject the null if their alpha = .05, and there were 13 participants in the sample.%3DQUESTION 8Click Save and Submit to save and submit. Click Save All Answers to save all answers.

The correct answer is C. The researcher would have rejected the null if their alpha =0.05 and there…

Answered: 10 for both samples QUESTION 7 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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Question 4 A company produces a range of devices for the mobile phone market. Its acceptance policy is that batches with more than 4% defective are returned to the supplier. A random sample of 150 plastic connectors reveals that 8 are defective. 95% Cl for the proportion defective, using one sample p, is (0.023304, 0.102382) What action should you take? O Accept the batch. Reject the batch. Wait and re-do the statistical test, the wrong statistical test was used.
You are testing to see if p1 = .2, p2 = .3 and p3 = .5. You take a sample of size 100 and tind 22 of the first population, 27 of the second population and 51 from the third population. Test Ho : P1 = .2, p2 = .3 and p3 = .5 against H1 : not Ho using alpha = .05.
Use the following sample data to answer the next questions. At the same time each day, a researcher records the temperature in each of three greenhouses. The table shows the temperatures in degrees Fahrenheit recorded for one week. Answer the ANOVA hypothesis steps below. (α= 0.025) Greenhouse 1 Greenhouse 2 Greenhouse 3 73 71 67 72 69 84 73 70 85 66 77 73 86 75 85 71 65 80 72 73 75 70 71 84 H0: Ha: F value: F-critical value: p-value: Conclusion: (Example answer: Reject or Fail to reject)
A sample of n=9 scores is obtained from a population with μ=45 and σ=6. If the sample mean corresponds to a z-score of 1.00, then what is the value of the sample mean? A. M=45 B. M=40 C. M=47 D. M=43
Given: σ = 0.9, δ = 3.25 − 3, α = 0.05 , Zα/2 = z0.025 What sample size would be required to detect a true mean, If we wanted the power to be at least 0.85?
The paragraph below is the study questions 16-21 d The Board of Transportation and Safety (BTS) wants to examine the safety mechanisms of A, B, and C-class cars. Samples of 7 cars for each of the classes were used. The board members collect the pressure exerted overall on the driver's body during a crash impact test. The BTS wants to see if there if a difference among the 3 classes of cars. Assume alpha of 0.05. What type of statistical test would be appropriate? O Pearson correlation O Multiple linear regression O one-way independent ANOVA O two-way independent ANOVA O simple linear regression QUESTION 17 Which of the following is the correct alternative hypothesis? O at least 2 of the means are different O all of the pairs of means are different O u1 = p2 = p3
A study was done to look at the relationship between number of lovers college students have had in their lifetimes and their GPAs. The results of the survey are shown below. Lovers 0 3 1 2 5 1 8 GPA 3.3 2.9 3.4 3.3 2.5 2.8 0.7 1. The p-value is: _____ (Round to four decimal places) 2. Use a level of significance of α=0.05 to state the conclusion of the hypothesis test in the context of the study (pick one). a. There is statistically significant evidence to conclude that there is a correlation between the number of lovers students have had in their lifetimes and their GPA. Thus, the regression line is useful. b. There is statistically significant evidence to conclude that a student who has had more lovers will have a lower GPA than a student who has had fewer lovers. c. There is statistically insignificant evidence to conclude that a student who has had more lovers will have a lower GPA than a student who has had fewer lovers. d. There is statistically insignificant…
1. Calculate r2 2. Interpret r2 in the context of this research. How large is this effect (small, medium, large)? 3. Make an statistical decision about the null using alpha =.05. Will you reject or retain the null based on your data? 4. Provide a conclusion of the data using Descriptive statistics and the context of the research.
The makers of Aain. 8.Gon burn gel claim that their gel witl relleve sunburn pain within 5 minutes of application. We wish to test this claim at the a 0.05 level of significance. A random sample of people with a sunburn were asked to rate their pain on a scale from 1-10 (with 1 Indicating very tittle pain, 10 indicating unbearable agony) and then apply the gel. Then the same group of people were asked to rate their pain on the same 10-point scale 5 minutes later. The data is given below. Let an be the mean pain indicator given by people before applying the gel and A be the mean pain indicator given by people 5 minutes after applying the gel. Assume that the pain differences have a normal distribution. (Round your results to three decimal places. Which would be correct hypotheses for this test? Ho B = PA, H: > HA Ha: B < HA, H:=HA %3D Ho:B HA, H:#B# HA %3D O Ho: HB HA, H1 : HB = HA Pain Indicator (on scale from 1-10): before gel after gel 9.5 3 8 6 4.5 7 2.5 8 5 6 6 6. 6. 7.5 4.5 10 9…
Two samples were drawn independently from two normal populations. Sample 1: 11.6, 11.35, 10.1, 10.85, 10.35, 10.6, and 11.1.Sample 2: 5.85, 8.85, 9.85, 11.85 and 7.85.Test to determine whether the means of the two population differs (use α=0.05).
1.3.2 Consider the output given below that was obtained using the One Proportion applet. Use information from the output to find the standardized statistic for a sample propor- tion value of 0.45. Probability of success (n): 0.30 Sample size (n): 25 Number of samples: 1000 O Animate Draw Samples Total = 1000 180 Mean = 0.301 SD = 0.091 120 60 0.08 0.16 0.24 0.32 0.40 0.48 0.56 0.64 Proportion of success
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