You recently concluded a paired t-test. Which of the following means that you found a 95% CI for µd of CI (-3.64, 1.23), and a p-value of .06.?
A.) The p-value is used to determine that we should not reject the null, because it is greater than .05, which shows there would be no difference between the means. Also, because 0 falls within our CI, 0 is a plausible value for µd and so we cannot reject the null.
B.) The p-value is used to determine that we should not reject the null, because it is greater than .05, which shows there would be no difference between the means. Also, because we did not reject the null, we do not need the CI.
C.) The p-value is used to reject the null, because it is greater than .05, which shows there would be no difference between the means. Also, because 0 falls within our CI, 0 is a plausible value for µd and so we reject the null.
Here, 0 falls within the CI for which suggest that the means are not significantly different. Also, since the p-value is greater than the alpha level of 0.05, we fail to reject the null hypothesis of no significant difference. Hence, our answer is: The p-value is used to determine that we should not reject the null, because it is greater than .05, which shows there would be no difference between the means. Also, because 0 falls within our CI, 0 is a plausible value for µd and so we cannot reject the null.
You recently concluded a paired t-test. Which of the following means that you found a 95%...
Suppose a sample of 49 paired differences that have been randomly selected from a normally distributed population of paired differences yields a sample mean d⎯⎯ =4.6d¯ =4.6 of and a sample standard deviation of sd = 7.6. (a) Calculate a 95 percent confidence interval for µd = µ1 – µ2. Can we be 95 percent confident that the difference between µ1 and µ2 is greater than 0? (Round your answers to 2 decimal places.) Confidence interval = [ , ] ;...
Suppose a sample of 49 paired differences that have been randomly selected from a normally distributed population of paired differences yields a sample mean d¯ =5.0d¯ =5.0 of and a sample standard deviation of sd = 7.8. (a) Calculate a 95 percent confidence interval for µd = µ1 – µ2. Can we be 95 percent confident that the difference between µ1 and µ2 is greater than 0? (Round your answers to 2 decimal places.) Confidence interval = [ , ] ;...
Suppose a sample of 49 paired differences that have been randomly selected from a normally distributed population of paired differences yields a sample mean d¯ =4.2 of and a sample standard deviation of sd = 7.6. (a) Calculate a 95 percent confidence interval for µd = µ1 – µ2. Can we be 95 percent confident that the difference between µ1 and µ2 is greater than 0? (Round your answers to 2 decimal places.) Confidence interval = [ ? , ?...
The MINITAB printout shows a test for the difference in two population means. Two-Sample T-Test and CI: Sample 1, Sample 2 Two-sample T for Sample 1 vs Sample 2 N Mean StDev SE Mean Sample 1 6 28.00 4.00 1.6 Sample 2 9 27.86 4.67 1.6 Difference = mu (Sample 1) - mu (Sample 2) Estimate for difference: 0.14 95% CI for difference: (-4.9, 5.2) T-Test of difference = 0 (vs not =): T-Value = 0.06 P-Value = 0.95...
You need to interpret the results of a recent paired t-test. What does the following mean? The test statistic t-value = -1.47 and the p-value = 0.01. A.) The small p-value means there is great chance of getting data like observed here again. Because the p-value is so small, the null hypothesis cannot be rejected. This indicates the data provided enough evidence to not reject the null. The test statistic t-value of -1.47 means the data is 1.47 standard errors...
1. You decide to test different samples for musical preferences: Baby Boomers, generation X and generation Z. Which type of inferential method, or case, is most appropriate for this scenario? A. ANOVA B. Independent samples C. Matched pairs 2.You have been asked to study the driving habits of marital couples; those living in the same household as a married couple. You want to measure the speed of spouse 1 and the speed of spouse 2 with the same vehicle. You...
Which statement is equivalent to a statistically significant independent t-test? Select all that are true. Retain the null hypothesis. The p-value is less than .05. The 95% confidence interval for the difference of means does not contain 0. Alpha is greater than .05.
Answer C is not correct. Which statement is not equivalent to the other three for a test comparing two sample means? Retain the null hypothesis. The p-value is greater than.05. The 95% confidence interval for the difference of means contains 0. The results are statistical significant
In a study of 795 randomly selectod medical malpractice lawsuits, t was found that 516 of them were dropped or dismiessed Use a 0 01 significance level to lest the clam thal most medical malpractice lawsuts are dropped or dismissed Which of the tollowing is the hypothesis test to be conducted? H p-05 O C. Ho p 05 111 p#05 E. Hop,05 H1 pc05 HI , p=05 OF. Ho P 05 H, p 05 What is the test statistic? Z-...
Consider a situation where we want to compare means, M1 and 42 of two populations, Group 1 and Group 2, respectively. A random sample of 40 observations was selected from each of the two populations. The following table shows the two-sample t test results at a = 5% assuming equal population variances: t-Test: Two-Sample Assuming Equal Variances Group 2 28652 33.460 40 Mean Variance Observations Pooled Variance Hypothesized Mean Difference d t Stat PTcut) one-tail Critical one-tail PTC-t) two-tail Critical...