To study the effectiveness of a weight-loss training method, a random sample of

Question

To study the effectiveness of a weight-loss training method, a random sample of 40 females went through the training. After a 2-month period of training, change of the weight is recorded for each participant. The researcher wants to test that the mean weight reduction is larger than 5 pounds at level 0.05. (a) The researcher knows that the population standard deviation is approximately 7 pounds. What is the required sample size to detect the power (1-) of 80% when the true weight reduction is 8 pounds? Is the sample size of 40 enough? (b) The average weight loss for the 40 women in the sample was 7.2 pounds with the sample standard deviation of 4.5 pounds. Is there sufficient evidence that the training is effective? Test using 0.05 (c) Find the power of the test conducted in part (b) if the true mean weight reduction is 8 pounds.

Explanation / Answer

(a) H0 : 5 pounds

Ha : < 5 pounds

Population standard deviation = 7 pounds

standard error of sample mean = / n = 7/ n

Here we will reject the null hypothesis if sample mean x > 5 + Z95% (/  n) = 5 + 1.645 * 7/ n = 5 + 11.515/ n

Here the power = 80%

so Power = Pr(x > 5 + Z95% (/  n) ; 8 ; 7/ n) = 0.80

Pr(x > 5 + 11.515/  n ; 8 ; 7/ n)

Pr(x < 5 + Z95% (/  n) ; 8 ; 7/ n) = 0.20

Z - value for p - value for 0.20

Z = -0.842 = (5 + 11.515/n - 8) /(7/ n)

3 = 11.515/n + 5.894/n

n = 5.803

n = 33.67

n = 34

soyes, sample size of 40 is enough.

(b) Average weight loss for the women = 40

sample mean x = 7.2 pounds

sample standard deviation s = 4.5 pounds

standard error of sample mean se0 = s/ sqrt(n) = 4.5/ 40 = 0.7115

Test statistic

t = (x - 5)/ se0 = (7.2 - 5)/ 0.7115 = 3.092

so here tcritical =t39,0.05,one-tailed = 1.6849

Here t > tcritical so we shall reject the null hypothesis and say that there is sufficient evidence that the training is effective.

(c) Here if the true mean weight reduction = 8 pounds.

so we shall reject the null hypothesis if sample mean > 5 + t39,0.05 se0

x > 5 + 1.6849 * 0.7115

x > 6.1988

so Power of the test when the test truely detect that it will reject the null hypothesis.

Pr(x > 6.1988 ; 8 ; 0.7115) = 1 - Pr(x < 6.1988 ; 8 ; 0.7115)

Z = (6.1988 - 8)/ 0.7115 = -2.53155

so Pr(x > 6.1988 ; 8 ; 0.7115) = 1 - Pr(Z < -2.53155) = 1 - 0.0057 = 0.9943

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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