When a survey asked subjects whether they would be willing to accept cuts in their standard of living to protect the environmen
1 answer:
Answer:
(a) The point estimate of the proportion of the population who would answer yes is 0.295.
(b) The margin of error for a 95% confidence interval is 0.0259.
(c) 95% confidence interval for population proportion is [0.2691 , 0.3209].
Step-by-step explanation:
We are given that a survey asked subjects whether they would be willing to accept cuts in their standard of living to protect the environment, 352 of 1195 subjects said yes.
Let = <u><em>sample proportion of subjects who said yes.</em></u>
(a) The point estimate of the proportion of the population who would answer yes = =
So, = <u>0.295</u>
(b) Margin of error is given by =
where, = level of significance = 1 - 0.95 = 5%
At 5% level of significance, z table gives critical value of 1.96 for two-sided interval.
Standard of error = =
= 0.0132
So, Margin of error for 95% confidence interval =
= <u>0.0259</u>
(c) 95% confidence interval for population proportion is given by =
Point estimate Margin of error
⇒ 0.295 0.0259
⇒ [0.295 - 0.0259 , 0.295 + 0.0259]
⇒ [0.2691 , 0.3209]
So, 95% confidence interval = [0.2691 , 0.3209]
The numbers in this interval represent that we are 95% confident that the population proportion will lie between 0.2691 and 0.3209.
(d) Assumptions needed for constructing a confidence interval are;
- The data must be sampled randomly.
- Sample values must be independent of each other.
- Data must follow normal distribution.
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Answer:
Step-by-step explanation:
Given that a small manufacturing firm has 250 employees. Fifty have been employed for less than 5 years and 125 have been with the company for over 10 years. So remaining 75 are between 5 and 10 years.
Suppose that one employee is selected at random from a list of the employees
A) Probability that the selected employee has been with the firm less than 5 years =
B) Probability that the selected employee has been with the firm between 5 and 10 years
=
C) Probability that the selected employee has been with the firm more than 10 years
=
a) P(A) = 0.2
P(C) = 0.5
P(A or B) = 0.2+0.3 = 0.5
P(A and C) = 0 (since A and C are disjoint)
Notice the grid, the points are (-5, -2) and (4, -1), thus
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~{{ 4}} &,&{{ -1}}~) % (c,d) &&(~{{ -5}} &,&{{ -2}}~) \end{array}~~ % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ d=\sqrt{[-5-4]^2+[-2-(-1)]^2}\implies d=\sqrt{(-5-4)^2+(-2+1)^2} \\\\\\ d=\sqrt{(-9)^2+(-1)^2}\implies d=\sqrt{81+1}\implies d=\sqrt{82}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%7B%7B%204%7D%7D%20%26%2C%26%7B%7B%20-1%7D%7D~%29%20%0A%25%20%20%28c%2Cd%29%0A%26%26%28~%7B%7B%20-5%7D%7D%20%26%2C%26%7B%7B%20-2%7D%7D~%29%0A%5Cend%7Barray%7D~~%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B-5-4%5D%5E2%2B%5B-2-%28-1%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-5-4%29%5E2%2B%28-2%2B1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-9%29%5E2%2B%28-1%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B81%2B1%7D%5Cimplies%20d%3D%5Csqrt%7B82%7D)