Answer :
Answer:
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.64})^2}=747.11[/tex]
And rounded up we have that n=748
Step-by-step explanation:
The confidence level is 90% and the significance level would be [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2 =0.05[/tex] and the critical value for this case is:
[tex] z_{\alpha/2}= 1.64[/tex]
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
And on this case we want a margin of error of [tex]ME =\pm 0.03[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
Since we don't hace any prior info for the population proportion we can use [tex]\hat p =0.5[/tex]. And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.64})^2}=747.11[/tex]
And rounded up we have that n=748
Using the sample size formula, the required number of sample is 752.
Using the confidence interval relation :
- [tex] n = \frac{pq}{(\frac{E}{Z_{crit}})^{2}} [/tex]
- p =sample proportion = 0.5
- q = 1 - p = 1 - 0.5 = 0.5
- Zcrit at 90% = 1.645
- E = Error Margin = 3% = 0.03
Substituting the values into the equation :
n = [tex] \frac{0.5 \times 0.5}{(\frac{0.03}{1.645})^{2} }[/tex]
[tex] n = \frac{0.25}{0.0003325}[/tex]
[tex] n = 751.67[/tex]
Therefore, the required number of sample is 752.
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