Respuesta :
Answer:
[tex] ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75[/tex]
±0.75 ounce
Step-by-step explanation:
Assuming this complete question: The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in that state. To do this, they have selected a random sample of 300 people from the designated population. The following results were recorded: xbar=34.5 ounces, s=7.9 ounces Given this information, if the leaders wish to estimate the mean milk consumption with 90 percent confidence, what is the approximate margin of error in the estimate?
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=34.5[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=7.9 represent the sample standard deviation
n=300 represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=300-1=299[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,199)".And we see that [tex]t_{\alpha/2}=1.653[/tex]
And the margin of error is given by:
[tex] ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75[/tex]
±0.75 ounce
Answer:
The approximate margin of error in the estimate is ±0.75 ounces.
Step-by-step explanation:
The question is incomplete:
The following results were recorded: xbar=34.5 ounces, s=7.9 ounces.
The sample size is n=300.
We will use the sample standard deviation to estimate the population standard deviation, so we will use the t-statistic.
To develop a confidence interval, we first have to calculate the degrees of freedom, and then look up in a t-students distribution table the critical value for a 90% confidence interval.
The degrees of freedom are:
[tex]df=n-1=300-1=299[/tex]
The critical value for a 90% CI is t=1.65.
Now, we can calculate the margin of error of the confidence interval as:
[tex]E=t\cdot s/\sqrt{n}=1.65*7.9/\sqrt{300}=13.035/17.32=0.75[/tex]
The lower and upper bounds of the confidence interval will be:
[tex]LL=\bar x-t\cdot s/\sqrt{n}=34.5-0.75=33.75\\\\UL=\bar x+t\cdot s/\sqrt{n}=34.5+0.75=35.25[/tex]
The confidence interval is (33.75, 35.25)
