A rubber ball has a radius of about 2.86 in. Can the ball be packaged in a box shaped like a cube with a volume of 125 in3?
1 answer:
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Answer:
0.1790
0.0738 , 0.5800, 0.7040
354
Step-by-step explanation:
a. Given:
n = 162
x = 29
c = 95%
The point estimate of the population proportion is the sample proportion. The sample proportion is the number of successes divided by the sample size:
p = x/n
= 29/162
= 0.1790
b. Given:
n=162
x = 110
c = 95%
The point estimate of the population proportion is the sample proportion. The sample proportion is the number of successes divided by the sample size:
p = x/n
=110/162
=0.6790
For confidence level 1 – = 0.95, determine z_
/2 = z_0.025 using table 1 (look up 0.025 in the table, the z-score is then the found z-score with opposite sign):
z_/2 = 1.96
The margin of error is then:
E = z_/2*√p(1-p)/n =1.96* √0.6790(1-0.6790)/162 =0.0738
The boundaries of the confidence interval are then:
p-z_/2*√p(1-p)/n = 0.6790-1.645√ 0.6790(1- 0.6790)/162 = 0.5800
p+z_/2*√p(1-p)/n = 0.6790+1.645√ 0.6790(1- 0.6790)/162 = 0.7040
c. Given:
n = 162
x = 110
c = 95%
The point estimate of the population proportion is the sample proportion. The sample proportion is the number of successes divided by the sample size:
p =x/n
=0.6790
Formula sample size:
p known: n = [z_/2 ]^2*pq/E^2
= [z_/2 ]^2*p(1-p)/E^2
n = [z_/2 ]^2*0.25/E^2
For confidence level 1 –= 0.95, determine z_
/2 = z_o.025 using table 1 (look up 0.025 in the table, the z-score is then the found z-score with opposite sign):
z_/2 = 1.96
p is known, then the sample size is (round up!):
n =[z_/2 ]^2*p(1-p)/E^2
= 354