Question

Assignment Questions Answer each question completely and send your work back to your instructor. Show all work. Justify your answers clearly and completely with formulas and numerical solutions. Showing your work will allow your instructor to give you partial credit if you make a mistake but show that you still understand the problem. 1. In a survey, 600 mothers and fathers were asked about the importance of sports for boys and girls. Of the parents interviewed, 70% said the genders are equal and should have equal opportunities to participate in sports. A. What are the mean, standard deviation, and shape of the distribution of the sample proportion of parents who say the genders are equal and should have equal opportunities? Be sure to justify your answer for the shape of the B. Using the norm approximation without the continuity correction, sketch the probability distribution curve for the distribution of p. Shade equal areas on both sides of the mean to show an area that represents a probability of .95, and label the upper and lower bounds of the shaded area as values of p-hat (not z-scores). Show your calculations for the upper and lower bounds. (2 points) C. Considering the sketch in part B, the shaded area shows a .95 probability of what happening? In other words, what does the probability of .95 represent? (2 points) D. Using the normal approximation, what's the probability a randomly drawn sample of parents of size 600 will have a sample proportion between 67% and 73%? Draw a sketch of the probability curve, shade the area representing the probability you're finding, and label the z-scores that represent the upper and lower bounds of the probability you're finding. Don't use the continuity correction. (2 points) E. Now, use the exact binomial calculation to find the probability of getting between, but not including, 67% and 73% of the respondents in a sample of 600 who say the genders are equal and should have equal opportunities. To use the exact binomial, you'll need to convert the proportions to counts by multiplying each proportion by 600. (1 point) F. Now try it again, but this time find the probability of getting at least 67% but no more than 73%. Use the exact binomial calculation. (1 point)

Answer 1

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Answer:

A)

For this distribution mean = np = 600*0.7 = 420

Standard Deviation = sqrt(npq) = aqrt(600*0.7*0.3) = 11.22

And the shape of the distribution is rightly skewed

B)

p = 0.7
n = 600

standard error, SE = sqrt(p(1-p)/n)
SE = sqrt(0.7(1-0.7) / 600) = 0.0187

Za/2 = Z0.025= 1.96

confidence interval:
p +/- [Za/2 * SE]
0.70 +/- 0.0367
( 0.6633 , 0.7367 )

lower bound = 0.6633

upper bound = 0.7367

p-hat is normally distributed with mean = p = 0.7, and standard deviation = 0.0187

951 0.6633 017 017367

c) In B. the shaded area shows probability 0.95 of that population proportion P lies in betweeen 0.663 and 0.737.

d) Here we need to find probability between 0.67 and 0.73

p ( 0.67 < P < 0.73 )

=

0.67 -0.70 0.73 – 0.70 р 0.70*(1-0.70) 600 0.70*(1-0.70) 600

= p ( -1.604 < z < 1.604 )

= p ( Z < 1.604 ) - p ( z < -1.604 )

= 0.9456 - 0.0544

= 0.8913

0.8913 0.67 Z=-1.604 0.73 z=1.604

E ) Here we need find exact probability ,

0.67 *600 = 402 , 0.73 *600 = 438

Here we have mean = 420 and standard deviation = 11.22

p ( 402 < x < 438 )

=

(402 – 420 P <P < 438 – 420 11.22 11.22

= p ( -1.604 < z < 1.604 )

= p ( Z < 1.604 ) - p ( z < -1.604 )

= 0.9456 - 0.0544

= 0.8913

F. at least 67%

p ( p $\geq$ 0.67 )

=

0.67 -0.70 0.70*(1-0.705 600

= p ( z $\geq$ -1.604)

= 1-p( z< -1.604)

=1-0.0544

= 0.9456

no more than 73%.

p ( p $\leq$ 0.73 )

=

0.73 – 0.70 0.70*(1-0.70) 600

= p ( z $\leq$ 1.604 )

= 0.9455