Respuesta :
Answer:
b) The margin of error for our 99% confidence interval would decrease.
True.
[tex] ME_i = z_{\alpha/2} \frac{\sigma}{\sqrt{64}}= \frac{z_{\alpha/2} \sigma}{8}[/tex]
And for the new sample size we have:
[tex] ME_f = z_{\alpha/2} \frac{\sigma}{\sqrt{100}}= \frac{z_{\alpha/2} \sigma}{10}[/tex]
As we can see the margin of error decrease
Explanation:
Assuming this problem: "You measure the lifetime of a random sample of 64 tires of a certain brand. In a follow-up study, more tires were available for testing, so you were able to measure the lifetimes of a random sample of 100 tires rather than 64. Which statement is TRUE?
a) The margin of error for our 99% confidence interval would increase.
b) The margin of error for our 99% confidence interval would decrease.
c) σ would decrease.
d) The margin of error for our 99% confidence interval would stay the same because the level of confidence has not changed."
For this case we want to find the margin of error from the experiment explained. On this case we just need to remember some concepts:
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean
s represent the sample standard deviation
[tex]\sigma[/tex] rrpreseent the population deviation
n represent the sample size
The margin of error for the sample mean would be given by:
[tex] ME=z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
With this we can analyze one by one the optiosn provided:
a) The margin of error for our 99% confidence interval would increase.
False. Assuming 99% for both intervals the value for [tex]z_{\alpha/2}[/tex] would be the same and the only factor that changes is the denominator.
The margin of error for the original sample size is:
[tex] ME_i = z_{\alpha/2} \frac{\sigma}{\sqrt{64}}= \frac{z_{\alpha/2} \sigma}{8}[/tex]
And for the new sample size we have:
[tex] ME_f = z_{\alpha/2} \frac{\sigma}{\sqrt{100}}= \frac{z_{\alpha/2} \sigma}{10}[/tex]
So we see that the margin of error decrease
b) The margin of error for our 99% confidence interval would decrease.
True.
[tex] ME_i = z_{\alpha/2} \frac{\sigma}{\sqrt{64}}= \frac{z_{\alpha/2} \sigma}{8}[/tex]
And for the new sample size we have:
[tex] ME_f = z_{\alpha/2} \frac{\sigma}{\sqrt{100}}= \frac{z_{\alpha/2} \sigma}{10}[/tex]
As we can see the margin of error decrease
c) σ would decrease.
False, the value for the population deviation is always fixed and not changes.
d) The margin of error for our 99% confidence interval would stay the same because the level of confidence has not changed.
False as we can see the margin of error changes, from the result above.
