Answer:
a) 95% of the confidence intervals created using this method would include the true proportion of shoppers who stock up on bargains.
b) Margin of error = 0.0270.
Step-by-step explanation:
a)
[tex]\hat{p}[/tex] = point estimate of p [tex]= 268/1020 =0.2627[/tex]
[tex]\hat{q}= 1 - 0.2627 = 0.7373[/tex]
[tex]SE = \sqrt{\hat{p}\hat{q}/n}\\\\=\sqrt{0.2627\times 0.7373/1020}=0.0138\\\alpha = 0.05[/tex]
From Table, critical values of [tex]Z= \pm1.96[/tex] ,
Lower limit
[tex]= \hat{p} - Z SE\\ = 0.2627 - (1.96 X 0.0138) \\ = 0.2627 - 0.0276 \\ = 0.236[/tex]
upper limit = 0.2627 + 0.0276
= 0.2903
Correct option:
95% of the confidence intervals created using this method would include the true proportion of shoppers who stock up on bargains.
b) Margin of error =Z SE
= 1.96 X 0.0141
= 0.0270.
