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Mrac [35]
3 years ago
8

12. If a manufacturer conducted a survey among randomly selected target market households and wanted to be 95% confident that th

e difference between the sample estimate and the actual market share for its new product was no more than 3%, what sample size would be needed?\
Mathematics
1 answer:
atroni [7]3 years ago
6 0

Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

And rounded up we have that n=1068

Step-by-step explanation:

We have the following info given:

Confidence= 0.95 the confidence level desired

ME =0.03 represent the margin of error desired

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

The confidence level is 95% or 0.95, the significance is \alpha=0.05 and the critical value for this case using the normal standard distribution would be z_{\alpha/2}=1.96

Since we don't have prior information we can use \hat p= 0.5 as an unbiased estimator

Also we know that ME =\pm 0.03 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

And rounded up we have that n=1068

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