Respuesta :
Answer:
We need n being at least 1600.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p, the mean is [tex]\mu = p[/tex] and the standard deviation is [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
The probability that a shopper will buy a packet of crackers after tasting the free sample is 0.200
This means that [tex]p = 0.2[/tex]
How large should n be so that the standard deviation of is no more than 0.01?
s at most 0.01, so a sample size of at least n when s = 0.01
[tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
[tex]0.01 = \sqrt{\frac{0.2*0.8}{n}}[/tex]
[tex]0.01\sqrt{n} = 0.4[/tex]
[tex]\sqrt{n} = \frac{0.4}{0.01}[/tex]
[tex]\sqrt{n} = 40[/tex]
[tex](\sqrt{n})^{2} = 40^{2}[/tex]
[tex]n = 1600[/tex]
We need n being at least 1600.
