Answer:
The upper bound of the 90% confidence interval is 0.181
Step-by-step explanation:
From the question we are told that
The sample size is n = 582
The number of accidents is k = 91
Generally the sample proportion is mathematically represented as
[tex]\r p = \frac{ k}{n} = \frac{582}{91 }[/tex]
=> [tex]\r p = 0.1564[/tex]
Generally the confidence level is 90% , the level of significance is [tex]\alpha = (100-90)\%[/tex]
=> [tex]\alpha = 0.10[/tex]
From the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is [tex]Z_{\frac{\alpha }{2} } = 1.645[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r p (1 - \r p)}{n} }[/tex]
=> [tex]E = 1.645* \sqrt{\frac{ 0.1564 (1 - 0.1564)}{582} }[/tex]
=> [tex]E = 0.0248[/tex]
Generally the 90% confidence interval is mathematically represented as
[tex]0.1564 - 0.0248 < p < 0.1564 -0.0248[/tex]
=> [tex]0.1316 < p < 0.1812[/tex]
Answer and Step-by-step explanation:
Random sample = 582
No. of accidents = 91
P = 91 / 582 = 0.156
Confidence interval for proportion of accidents:
P ± z √(p(1-p)/n)
Where,
n = 582
p = 0.156
z = 1.645
Formula:
P ± z √(p(1-p)/n)
put these values in formula, we get:
= 0.156 ± 1.645 √(0.156(1-0.156)/582)
= 0.156 ± 1.645 √(0.1316/582)
= 0.156 ± 1.645 (0.01483)
= 0.156 ± 0.0244
= (0.1804, 0.1316)
