A measure of variability provides information about a. Level of performance b. S

Question

A measure of variability provides information about a. Level of performance b. Shape of the distribution c. The spread of the distribution A sample has a M = 39.5 and an s = 4.3. In a 2 tailed hypothesis test with a level of 05, this sample produces a t statistic of 2.14. Based on this information, the correct statistical decision is a. Reject the null hypothesis b. Do not reject the null hypothesis c. Cannot answer the question without additional information An independent-measures t test uses one sample with n = 10 and a 2^nd sample of n = 15, the t statistic from this experiment will have degrees of freedom of a. 23 b. 24 c. 25 d. Cannot answer without additional information A sample of n = 25 individuals is selected from a population with mu =80 and a treatment is administered to the sample. If the treatment has a large effect, then a. The sample mean should be very different from 80 and should lead you to rejct the null hypothesis b. The sample mean should be very different from 80 and should lead you to not reject the null hypothesis c. The sample mean should be very close to 80 and should lead you to reject the null hypothesis d. The sample mean should be very close to 80 and should lead you to not reject the null hypothesis A distribution has a mean of 60 and a sigma = 8. For a score of 72, the Z score is a. +12 b. -12 C. +1.5 d. -1.5

Explanation / Answer

11) option C

12)as we do not know degree of freedom;

option C

13)degree of freedom =n1+n2-2=10+15-2=23

option A

14)option A

15) a score =(72-60)/8=1.5

option C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.