Respuesta :
Answer:
a) By the central limit theorem we know that if the random variable X="quantitative scores for young women", and we know that this distribution is normal.
[tex]X \sim N(\mu, \sigma=60)[/tex]
And we are interested on the distribution for the sample mean [tex]\bar X[/tex], we know that distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And with that we have the assumptions required to apply the normal distribution to create the confidence interval since the distribution for [tex]\bar X[/tex] is normal.
b) [tex]270.283\leq \mu \leq 279.717[/tex]
c) We are confident (99%) that the true mean for the quantitative scores for young women is between (270.283;3279.717)
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
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=275[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=60[/tex] represent the population standard deviation
n=1077 represent the sample size
Part a
By the central limit theorem we know that if the random variable X="quantitative scores for young women", and we know that this distribution is normal.
[tex]X \sim N(\mu, \sigma=60)[/tex]
And we are interested on the distribution for the sample mean [tex]\bar X[/tex], we know that distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And with that we have the assumptions required to apply the normal distribution to create the confidence interval since the distribution for [tex]\bar X[/tex] is normal.
Part b
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that [tex]z_{\alpha/2}=2.58[/tex]
Now we have everything in order to replace into formula (1):
[tex]275-2.58\frac{60}{\sqrt{1077}}=270.283[/tex]
[tex]275+2.58\frac{60}{\sqrt{1077}}=279.717[/tex]
So on this case the 99% confidence interval would be given by (270.283;3279.717)
[tex]270.283\leq \mu \leq 279.717[/tex]
Part c
We are confident (99%) that the true mean for the quantitative scores for young women is between (270.283;3279.717)