Week 3 Assignment

Week 3 Assignment
7.11

A. The shape should be approximately shape because the central intend theorem. The sample size could be considered epic so the sampling distribution will be normal.

B. The pie-eyed is the same as the population mean, 20.
The standard deviation is 4/sqrt(64) = 4/8 = 0.5

C. z = (xbar - mu)/(sigma/?n)
z = (21 - 20)/(4/?64)
z = 1 / 0.5
z = 2
p = 0.9772
p = 1 - 0.9772 = 0.0228

D. z = (xbar - mu)/(sigma/?n)
z = (19.385 - 20)/(4/?64)
z = -0.615 / 0.5
z = -1.23
p = 0.1093

7.30

A. z = (phat - p)/sqrt[p (1-p)/n]
z = (0.32 - 0.3) / sqrt [0.3(1 - 0.3)/1011]
z = 0.02 / sqrt (0.00020772)
z = 1.421

p = 0.9223

Since we want greater than, p = 1 - 0.9223 = 0.0777

B. Maybe, but we did not set an Alpha level sooner beginning.

8.8
A. if ? = 0.05 then CI (95%) for the mean is 5.46±z(0.025)2.47/?100 = (4.976,5.944)
if ? = 0.



01 then CI (99%) for the mean is 5.46±z(0.005)2.47/?100 = (4.824,6.096)

B. Yes, because the upper limit (5.944) < 6

C. No, because the upper limit (6.096) >6
D. We are 95% confident that the mean is less than 6
8.38 95% confidence interval:

p +/- z * sqrt [p(1 - p)/n]
0.5571 +/- 1.96 * sqrt [( 0.5571 * 0.4429)/350]
0.5571 +/- 1.96 * sqrt [0.00070496]
0.5571 +/- 0.05204
(0.5051, 0.6092)

Since 0.48 is not at bottom the confidence interval, we can be 95% certain that the true(a) proportion is above 0.48If you want to get a wide of the mark essay, order it on our website: Orderessay



If you want to get a full essay, wisit our page: write my essay .