For a set alpha level of .05, a predictor that has a p-value less than or equal to .05 suggests that variations in the predictor's value are related to changes in the outcome (response) variable. True or False
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Correlation
Correlation defines a relationship between two independent variables. It tells the degree to which variables move in relation to each other. When two sets of data are related to each other, there is a correlation between them.
Linear Correlation
A correlation is used to determine the relationships between numerical and categorical variables. In other words, it is an indicator of how things are connected to one another. The correlation analysis is the study of how variables are related.
Regression Analysis
Regression analysis is a statistical method in which it estimates the relationship between a dependent variable and one or more independent variable. In simple terms dependent variable is called as outcome variable and independent variable is called as predictors. Regression analysis is one of the methods to find the trends in data. The independent variable used in Regression analysis is named Predictor variable. It offers data of an associated dependent variable regarding a particular outcome.
For a set alpha level of .05, a predictor that has a p-value less than or equal to .05 suggests that variations in the predictor's value are related to changes in the outcome (response) variable.
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- Twenty laboratory mice were randomly divided into two groups of 10. Each group was fed according to a prescribed diet. At the end of 3 weeks, the weight gained by each animal was recorded. Do the data in the following table justify the conclusion that the mean weight gained on diet B was greater than the mean weight gained on diet A, at the ? = 0.05 level of significance? Assume normality. (Use Diet B - Diet A.) Diet A 11 8 14 6 13 9 13 6 11 13 Diet B 18 11 7 13 15 12 21 16 19 12 (a) Find t. (Give your answer correct to two decimal places.)(ii) Find the p-value. (Give your answer correct to four decimal places.)(b) State the appropriate conclusion. Reject the null hypothesis, there is significant evidence that diet B had a greater weight gain.Reject the null hypothesis, there is not significant evidence that diet B had a greater weight gain. Fail to reject the null hypothesis, there is significant evidence that diet B had a greater weight gain.Fail to reject the null…untreated depression will lose at least one friendship while not receiving treatment. On the other hand, when people with depression are treated with medication and counseling, only 2 out of 100 will lose at least one friendship during treatment. The absolute risk of losing at least one friendship while dealing with untreated depression is Answer: 25 / 100 = 0.25 = 25%The 2006 Statistical Abstract of the United States reports on a survey that asked a national sample of 80,000 American households about pet ownership. Suppose (for now) that one-third of all American households own a pet cat. This survey found that 32.8% of the households sampled owned a pet cat. Is this number a parameter or a statistic? Calculate the Z-score of this sample proportion (0.328).
- Calculate SSTrA study was performed to determine the percentage of people who wear life vests while out on the water. A researcher believed that the percentage was different for those who rode jet skis compared to those who were in boats. Out of 500500 randomly selected people who rode a jet ski, 85%85% wore life vests. Out of 250250 randomly selected boaters, 90.4%90.4% wore life vests. Using a 0.050.05 level of significance, test the claim that the proportion of people who wear life vests while riding a jet ski is not the same as the proportion of people who wear life vests while riding in a boat. Let jet skiers be Population 1 and let boaters be Population 2. Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below. H0Ha: p1=p2: p1⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯p2 Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places. 3) draw a conclusion and form a decision (word it like "we fail to reject the null hypothesis and…The variance (5x) corresponds with which of the following ANOVA terms? a. mean square b. treatment effect O c. factor d. sum of squares
- A clinical psychologist is investigating the relationship between sleep and feelings ofanxiety. For a sample of 15 individuals, he asks each participant to indicate how many hours theytypically sleep each night and each participant also completes an anxiety assessment. The followingdata is obtained:Hours of sleep: M=6.0 SS=16.0Anxiety scores: M=8.0 SS=64.0SP = - 20.0 a. Compute the correlation between hours of sleep and anxiety scores.Historically, Olympic officials have reported testosterone levels in general should be u = 538 ng/dL, with o = 189, with x being normally distributed. Summer Olympic sprinters (n = 8) from the U.S. were tested for testosterone levels the morning after the opening ceremonies. Their sample mean was ng/dL. Do these data indicate that the sprinters had an overall average testosterone level greater than 538? Use a = .05. State a, H,, H1, p-value, rejection, or non-rejection, and be certain to interpret your results.A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 54 men and 75 women to participate in the study. Each subject was required to step up and down a 6-inch platform. The pulse of each subject was then recorded. The following results were obtained. N Mean StDev SE Mean Men Women 95% Cl for mu Men - mu Women - 9.81, - 0.79) T-Test mu Men = mu Women (vs H2 Oc. Họ: H1 = H2; Hai H1A survey of the population of 51 students in Psych 311 asks how many credits students are taking and obtains the following normally-distributed results: ∑ X= 686 and ∑X2= 9832.92. Using this information, answer the following three questions. What is the percentile rank of a student who is enrolled in 15 credits? (Hint: This is the same thing as asking “what percentage of students is enrolled in 15 credits or less?”) MUST SHOW WORK FOR THIS PROBLEM What percentage of students are enrolled in between 12 and 18 credits? (MUST SHOW WORK FOR THIS PROBLEM) What is the 25th percentile? (Hint: This is the same as asking what score separates the bottom 25% of the distribution from the top 75%.) MUST SHOW WORK FOR THIS PROBLEMAccording to previous studies, the mean distance each visitor in Greenspan National Park hikes during their visit is 30 kilometers. The park recently closed its shuttle system, which used to transport hikers to many of the park's most popular hiking trails. Because of this, an administrator at the park suspects the mean distance, u, is now less than 30 kilometers. The administrator chooses a random sample of 45 visitors. The mean distance hiked for the sample is 27.2 kilometers. Assume the population standard deviation is 9.9 kilometers. Can the administrator conclude that the mean distance hiked by each visitor is now less than 30 kilometers? Perform a hypothesis test, using the 0.10 level of significance. (a) State the null hypothesis H, and the alternative hYpothesis H,. OAccording to previous studies, the mean distance each visitor in Greenspan National Park hikes during their visit is 21 kilometers. The park recently closed its shuttle system, which used to transport hikers to many of the park's most popular hiking trails. Because of this, an administrator at the park suspects the mean distance, u, is now less than 21 kilometers. The administrator chooses a random sample of 85 visitors. The mean distance hiked for the sample is 20.5 kilometers. Assume the population standard deviation is 8.3 kilometers. Can the administrator conclude that the mean distance hiked by each visitor is now less than 21 kilometers? Perform a hypothesis test, using the 0.10 level of significance. (a) State the null hypothesis H and the alternative hypothesis H . H: 0 OSEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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