Suppose that, for a t-test, your computed value for t is +3.28. The critical value of t is +2.048. Explain what this means. Do you reject the null hypothesis or not? Now suppose that you have 28 degrees of freedom and are using a two-tailed (nondirectional) test. Draw a simple figure to illustrate the relationship between the critical and the computed values of t for this result.
Our computed value for t is larger than the critical value. This means that the groups are different. The two means are different from each other. Since it is different then the null hypothesis is rejected. The critical value is the baseline, above which the null hypothesis must be rejected.
Please find the figure attached
Suppose that, for a t-test, your computed value for t is +3.28. The critical value of...
Calculate the critical degrees of freedom and identify the critical t value for a single-sample t test in each of the following situations, using p=.05 for all scenarios. Then, state whether the null hypothesis would be accepted or rejected: 10) Two-tailed test, N = 10, t = 2.35 df= (answer) critical t = (answer) Accept or Reject Ho: (answer)
Rebecca is using a t-distribution for a hypothesis test. She finds that the critical value is 3.64 and the test statistic is 3.95. If this is a Right Tailed Test, what conclusion should she draw? Reject the Null Hypothesis Reject the Hypothesis Test Reject the Alternative Hypothesis Do Not take any action
Do a two-sample test for equality of means assuming unequal variances. Calculate the p-value using Excel. (a-1) Comparison of GPA for randomly chosen college juniors and seniors: x⎯⎯1x1 = 4.75, s1 = .20, n1 = 15, x⎯⎯2x2 = 5.18, s2 = .30, n2 = 15, α = .025, left-tailed test. (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick"...
Do a two-sample test for equality of means assuming unequal variances. Calculate the p-value using Excel. (a-1) Comparison of GPA for randomly chosen college juniors and seniors: x⎯⎯1x1 = 4.75, s1 = .20, n1 = 15, x⎯⎯2x2 = 5.18, s2 = .30, n2 = 15, α = .025, left-tailed test. (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick"...
Do a two-sample test for equality of means assuming unequal variances. Calculate the p-value using Excel. (a-1) Comparison of GPA for randomly chosen college juniors and seniors: x⎯⎯1x1 = 4.75, s1 = .20, n1 = 15, x⎯⎯2x2 = 5.18, s2 = .30, n2 = 15, α = .025, left-tailed test. (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick"...
Do a two-sample test for equality of means assuming unequal variances. Calculate the p-value using Excel. (a-1) Comparison of GPA for randomly chosen college juniors and seniors: x⎯⎯1 = 4, s1 = .20, n1 = 15, x⎯⎯2 = 4.25, s2 = .30, n2 = 15, α = .025, left-tailed test. (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick"...
Do a two-sample test for equality of means assuming unequal variances. Calculate the p-value using Excel. (a-1) Comparison of GPA for randomly chosen college juniors and seniors: x⎯⎯1 = 4, s1 = .20, n1 = 15, x⎯⎯2 = 4.25, s2 = .30, n2 = 15, α = .025, left-tailed test. (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick"...
Question 11 1 pts In 2012, the mean weight of women was reported at 2012 = 162 lbs. Suppose that an investigator hypothesizes that weights are even higher in 2015. She collected a sample of 29 women in the year of 2015, from which she measured a mean weight of 158 lbs with a standard deviation of 10 lbs. An two-tailed alpha level = .05 is applied. Select ALL correct statement(s) below: The investigator aims to compare a sample mean...
I spefically need to see how the test statistic and critical value is calculated. Test the claim that the proportion of men who own cats is significantly different than 80% at the 0.02 significance level. The null and alternative hypothesis would be: The test is: left-tailed right-tailed two-tailed Based on a sample of 55 people, 78% owned cats The test statistic is: (to 2 decimals) The positive critical value is: (to 2 decimals) Based on this we: Reject the null...
Question 11 3 pts For this problem you will need a t-table. What is the t-value needed for the critical value for a two- tailed hypothesis test with degrees of freedom = 15 and alpha = 10? +/- 1.753 +/- 2.624 +/- 1.341 +/- 1.761