Introduction to the t Statistic Graded Assignment Read Chapter 9 Back to Assignm

Question

Introduction to the t Statistic Graded Assignment Read Chapter 9 Back to Assignment Due Sunday 11.12.17 at 1145 PM Keep the Highest: :5 5. Two-tailed hypothesis testing Step by step Aa S-denosyt menore (SAM-e) is a naturaty occurring compound i, hunan that is thought to have an emed on depression symptoms Suppose that a researcher is interested in testing SAM-e on patients who are struggling with cancer. He obtains a sample of m- 25 patients and asks each person to take the suggested dosage each day for 4 weeks. At the end of the 4-week period, each individual takes the Beck Depression Inventory (BDr), which is a 21-tem, multiple-choice self-report inventory for measuring the severilty of depression. The scores from the sample produced a mean of M-26.3 with standard deviation of s 1.01. In the general pplaton, of aror patents, the standardeed test is known to havepopulation mean of p 27.1. Becate there ane no previous studies using SAM-e wilth this population, the researcher doesnt know how it will affect these patients; therefore, he uses a two-taled single-sample t test to test the hypothesis From the folowing, select the correct nul and aternative hypotheses for this study Assume that the depression scores among patients taking SAM-e are normalhy distributed. You w first need to determine the degrees of freedom. There are degrees of freedom. Use the t distribution table to find the critical region for -05 The critical t scores (the values that define the borders of the critical region) are The estimated standard error is The t statistic is The t statistic in the critical region. Therefore, the nultl hypothesis rejected Therefore, the researcher condude that SAM-e has a significant effect on the moods of cancer patients.

Explanation / Answer

Performing t-test here because sample size is less than 30.

Mean could larger or smaller than the hypothesized value 27.1 so we have to conduct 2 tail t test.( 4th option)

t-satistic = (26.3 - 27.1)/ (1.01/sq.root(26)) = -4.04 with degrees of freedom = n-1 = 26-1 = 25

using t distribution table and significance of 0.05 cutoff value is -2.06 so we can safely say that our t statistic doesn't lie in the rejection region.

So we cannot reject the null hypothesis and thus results are not significant yet with 5% significance level

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