Question

In a study in Scotland (as reported by Devlin 2009), researchers left a total of 320 wallets around Edinburgh, as though the wallerts were lost. Each contained contact information including an address. Of the wallets, 146 were returned by the people who found them. With the following steps, use the data to estimate the proportion of lost wallets that are returned, and give a 95% confidence interval for this estimate.
Required:
a. What is the observed proportion of wallets that were returned (rounded to the nearest thousandths)?
b. Calculate p to use in the Agresti-Coull method of calculating a 95% confidence interval for the population proportion rounded to the nearest thousandths).
c. Calculate the lower bound of the 95% confidence interval (rounded to the nearest thousandths)
d. Calculate the upper bound of the 95% confidence interval (rounded to the nearest thousandths)

Answer 1

Answer:

a) The observed proportion of wallets that were returned

  p = 0.45625

b) 95% of confidence intervals for Population proportion

 0.40168  , 0.51082)

c) The lower bound of the 95% confidence interval = 0.40168

d) The upper bound of the 95% confidence interval = 0.51082

Step-by-step explanation:

Step(i):-

a)

Given data the  researchers left a total of 320 wallets around Edinburgh, as though the wallets were lost. Each contained contact information including an address. Of the wallets, 146 were returned by the people who found them

Given sample size 'n' = 320

  Given data          'x ' = 146

Sample proportion

              $p = \frac{x}{n}$

             $p = \frac{x}{n} = \frac{146}{320} = 0.45625$

Step(ii):-

b) 95% of confidence intervals for Population proportion

Level of significance = 95% or 0.05%

$Z_{\frac{\alpha }{2} } = Z_{\frac{0.05}{2} } = Z_{0.025} = 1.96$

95% of confidence intervals for Population proportion are determined by

$(p - Z_{0.025} \frac{\sqrt{p(1-p)} }{\sqrt{n} } , p + Z_{0.025} \frac{\sqrt{p(1-p)} }{\sqrt{n} })$

$(0.45625 - 1.96\frac{\sqrt{0.45625(1-0.45625)} }{\sqrt{320} } , 0.45625 + 1.96\frac{\sqrt{0.45625(1-0.45625)} }{\sqrt{320} })$

(0.45625 - 0.05457 , 0.45625 + 0.05457)

(   0.40168  , 0.51082)

c) The lower bound of the 95% confidence interval = 0.40168

d) The upper bound of the 95% confidence interval = 0.51082