Question

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. n = 195, x = 162; 95% confidence

Answer 1

Given:
n = 195, sample size.
x = 162, successes in the sample

The proportion is
p = x/n = 162/195 = 0.8308

n* p = 195*0.8308 = 162
n*(1-p) = 195*(1 - 0.8308) = 33
If n*p >= 10, and n*(1-p) >= 10, then the sample proportions will have a normal distribution. This condition is satisfied.

The proportion mean is
μ = 0.8308
The proportion standard deviation is
$\sigma = \sqrt{ \frac{p(1-p)}{n} } = \sqrt{ \frac{0.8308(1-0.8308)}{195} } =0.0269$

σ/√n = 0.0269/√195 = 0.00192

At the 95% confidence level, the interval for the population proportion is
(μ - 1.96(σ/√n), μ + 1.96(σ/√n))
= (0.8308 - 1.96*0.00192, 0.8308 + 1.96*0.00192)
= (0.827, 0.8345)

Answer: The 95% confidence interval is (0.827, 0.835)

Answer 2

At 95% confidence level the interval for the population proportion is (0.827 , 0.8345)

Step-by-step explanation:

Given:

Sample size (n) = 195

Successes in sample (x) = 162

Calculation :

production is,

$p = \dfrac{x}{n} = \dfrac {162}{195} = 0.8308$

$\rm n \times p = 195 \times 0.8308 = 162 \;\;(approx)$

$\rm n\times (1-p) = 195(1-0.8308) = 33 \;(approx)$

if ,  $\rm n\times p \geq 10 \; and n(1-p)\geq 10$ ,

then the sample proportion have a normal distribution.

And the above condition is satisfied.

$\mu = .8308$

$\sigma = \sqrt{\dfrac{p(1-p)}{n}} = \sqrt{\dfrac {0.8308(1-0.8308)}{195}}= 0.0269$

$\dfrac{\sigma}{\sqrt{n} } = \dfrac{0.0269}{\sqrt{195} } = 0.00192$

Therefore, at 95% confidence level the interval for the population proportion is,

$(\mu-1.96(\dfrac{\sigma }{\sqrt{n} }), \mu + 1.96(\dfrac{\sigma }{\sqrt{n} })) = (0.8308-1.96\times 0.00192 \;, \; 0.8308+1.96\times 0.00192)$

$(\mu-1.96(\dfrac{\sigma }{\sqrt{n} }), \mu + 1.96(\dfrac{\sigma }{\sqrt{n} })) = (0.827 , 0.8345)$

At 95% confidence level the interval for the population proportion is (0.827 , 0.8345)

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