Sam wants to estimate the percentage of people who have a smart phone. She wants to create a 95% confidence interval which has a margin of error (or ME) of at most 2%. How many people should be polled to create the confidence interval?

Respuesta :

Answer: 2401

Step-by-step explanation:

We know that if prior proportion of population doesn't exist , then the formula to find the population proportion is given by :_

[tex]n=0.25(\dfrac{z_{\alpha/2}}{E})^2[/tex]

Given : Level of confidence = 0.95

Significance level : [tex]\alpha=1-0.95=0.05[/tex]

Critical value : [tex]z_{\alpha/2}=1.96[/tex]

Margin of error : [tex]E=0.02[/tex]

Then ,we have

[tex]\Rightarrow n=0.25(\dfrac{1.96}{0.02})^2=2401[/tex]

Hence, the minimum sample size needed =2401

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