Answer:
No for n = 50; Yes for n = 100
Step-by-step explanation:
A researcher is studying a population in which 57% of the population has a certain trait. The researcher wants a sampling distribution with a standard deviation that is less than 6%.
To determine the standard deviations for different sample sizes, we can use the formula for the standard deviation of the sampling distribution of sample proportions:
[tex]\sigma_{\hat{p}} = \sqrt{\dfrac{p(1-p)}{n}}[/tex]
where:
- p is the population proportion.
- n is the sample size.
In this case p = 0.57, so:
[tex]\sigma_{\hat{p}} = \sqrt{\dfrac{0.57(1-0.57)}{n}}\\\\\\\\\sigma_{\hat{p}} = \sqrt{\dfrac{0.2451}{n}}[/tex]
Now, we can substitute the different sample sizes (n) into the formula and calculate the corresponding standard deviations:
When n = 50:
[tex]\sigma_{\hat{p}} = \sqrt{\dfrac{0.2451}{50}}=0.07001428...\approx7.0\%[/tex]
When n = 100:
[tex]\sigma_{\hat{p}} = \sqrt{\dfrac{0.2451}{100}}=0.0495075...\approx 5.0\%[/tex]
Therefore:
- For n = 50, the standard deviation is approximately 7.0%.
- For n = 100, the standard deviation is approximately 5.0%.
So, a sample size of 50 would not give a standard deviation that is less than 6%, however, a sample size of 100 would indeed provide a standard deviation that is less than 6%:
[tex]\Large\boxed{\boxed{\textsf{No for $n = 50$; Yes for $n = 100$}}}[/tex]
