Stat 1-Online-Fall 2017-Prof. Barbee tymor abolson 1 1/22/17 3:20 PM Homework: 9

Question

Stat 1-Online-Fall 2017-Prof. Barbee tymor abolson 1 1/22/17 3:20 PM Homework: 9.3- Homework Save Score: 0.67 of 1 pt 20f 7 (4 complete) HW Score: 52.38%, 3.67 of 7 pt Instructor-created question Question Help Treatment Placebo independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.05 significance level for both parts. A study was done using a treatment group and a placebo group. The results are shown in the table. Assume that the two samples are1H2 31 28 x2.36 0.93 2.67 0.54 H1 : 1 The test statistic, t, is©(Round to two decimal places as needed.) Use the critical value or p-value to determine the result of this problem. There is no where to enter your critical value or p-vlaue but you need them to find a conclusion State the conclusion for the test A. Fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean. O B. Reject the null hypothesis. There is sullicient evidence to warrant rejection of the claim that the two samples are from populations with the same mean OC. Fai to reject the null hypothesis. There is sulficent evidence to warrant rejection of the claim that the two samples are from populations with the same mean. D. Reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean. b. Construct a confidence interval suitable for testing the claim that the two samples are from populations with the same mean. Click to select your answer(s) and then click Check Answer. All oarts showina

Explanation / Answer

The statistical software output for this problem is:

Two sample T summary hypothesis test:
1 : Mean of Population 1
2 : Mean of Population 2
1 - 2 : Difference between two means
H0 : 1 - 2 = 0
HA : 1 - 2 0
(with pooled variances)

Hypothesis test results:

The test statistic is the value of T - Stat in the table. Hence,

Test statistic = -1.58

Difference Sample Diff. Std. Err. DF T-Stat P-value 1 - 2 -0.31 0.19565154 57 -1.5844496 0.1186

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