Respuesta :
Answer:
a) point estimate of the proportion of items returned for the population of sales transactions at the store in the given city
Point estimate = 0.3
b) The 95% confidence interval for the proportion of returns at the store in the given city
( 0.19958,0.40042)
Step-by-step explanation:
Given the Population proportion 'P'= 16% = 0.16
Given data a department store in a certain city sampled 80 items sold in January and found that 24 of the items were returned
The sample proportion [tex]p = \frac{x}{n}[/tex]
here x =24 and n = 80
The sample proportion [tex]p = \frac{x}{n} = \frac{24}{80} = 0.3[/tex]
a) Point estimate:-
A point estimate of the true proportion P
Point estimate = [tex]p = \frac{x}{n}[/tex]
point estimate of the proportion of items returned for the population of sales transactions at the store in the given city
[tex]p = \frac{x}{n} = \frac{24}{80} = 0.3[/tex]
b) The 95% confidence interval for the proportion
The 95% confidence interval for the proportion of returns at the store in the given city
[tex](p - z_{\alpha } \sqrt{\frac{p(1-p)}{n} } , p + z_{\alpha }\sqrt{\frac{p(1-p)}{n} } )[/tex]
substitute all values we get ,
[tex](0.3 - 1.96 \sqrt{\frac{0.3(1-0.3)}{80} } , 0.3 + 1.96\sqrt{\frac{0.3(1-0.3)}{80} } )[/tex]
on simplification, we get
(0.3-0.10042 ,0.3+0.10042)
( 0.19958,0.40042)
Conclusion:-
a) point estimate of the proportion of items returned for the population of sales transactions at the store in the given city
Point estimate = 0.3
b) The 95% confidence interval for the proportion of returns at the store in the given city
( 0.19958,0.40042)
