Answer :
Answer:
Preference: Online
[tex]Mean = Median = Mode=\hat p_{O}=0.5625[/tex]
Preference: Mall
[tex]Mean = Median = Mode=\hat p_{M}=0.3611[/tex]
Step-by-step explanation:
The sample size is, n = 1008.
Number of people who prefer shopping online is, n (O) = 567.
Number of people who prefer shopping in a mall is, n (M) = 364.
Number of people who were not sure is, n (NS) = 77.
Compute the proportion of people who prefer shopping online as follows:
[tex]\hat p_{O}=\frac{n(O)}{n} =\frac{567}{1008}=0.5625[/tex]
Compute the proportion of people who prefer shopping in a mall as follows:
[tex]\hat p_{M}=\frac{n(M)}{n} =\frac{364}{1008}=0.3611[/tex]
⇒ According to the Central limit theorem, if the sample drawn from an unknown population, with mean μ and standard deviation σ, is large (n ≥ 30) then the sampling distribution of the sample proportion ([tex]\hat p[/tex]) follows a Normal distribution with mean [tex]\mu=\hat p[/tex] and standard deviation [tex]\sigma =\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex]
So the sampling distribution of the sample proportion of people who prefer shopping online [tex](\hat p_{O})[/tex] and the the sample proportion of people who prefer shopping in a mall [tex](\hat p_{M})[/tex] follows a normal distribution.
⇒ In case of a normal distribution the mean, median and mode are same.
Thus, the mean, median and mode of the sample proportion of people who prefer shopping online is:
[tex]Mean = Median = Mode=\hat p_{O}=0.5625[/tex]
And the mean, median and mode of the sample proportion of people who prefer shopping in a mall is:
[tex]Mean = Median = Mode=\hat p_{M}=0.3611[/tex]