Question

Giorgio surveyed 45 randomly selected shoppers at the mall and found that approximately 24% believe that the food court needs to be remodeled. To the nearest percent, with a confidence level of 99% (z*-score 2.58), what is the confidence interval for the proportion of shoppers who believe the food court needs to be remodeled? E = z* and C = + E

Answer 1

Between 8 and 40 is the correct answer i just took the test

Answer 2

Answer with explanation:

Number of People Surveyed (S)= 45

Confidence Level = 99%

Percentage of population who believe that food court needs to be remodeled =24 %

→So, 24% of 45

$=\frac{24}{100}\times 45=10.80$

= 11 people

Probability of population who thinks that food court needs to be remodeled(P)

        $=\frac{11}{45}$  

$Z_{99 percent}= 2.58$

Confidence level for a population is given by

  $=P \pm Z\sqrt{\frac{P*(1-P)}{S}}$

Substituting the values, of Z, P and S in above equation

    $=0.25\pm 2.58\sqrt{\frac{0.25*0.75}{45}}\\\\=0.25 \pm 2.58*0.064549\\\\=0.25 \pm 0.17\\\\= 0.25 - 0.17 , 0.25+0.17\\\\=0.08 , 0.42$

→→0.08 ≤proportion of shoppers who believe the food court needs to be remodeled≤0.42

→→8%≤proportion of shoppers who believe the food court needs to be remodeled≤42%