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Anna35 [415]
2 years ago
9

When Poll Pasked students to identify their favorite kind

Mathematics
1 answer:
Arada [10]2 years ago
3 0

Answer:

120 students

Step-by-step explanation:

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A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selec
VMariaS [17]

Answer:

19.1-3.355\frac{1.5}{\sqrt{9}}=17.42    

19.1+3.355\frac{1.5}{\sqrt{9}}=20.78    

And the best option would be:

C. [17.42,20.78]

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".

\bar X=19.1 represent the sample mean

\mu population mean (variable of interest)

s=1.5 represent the sample standard deviation

n=9 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=9-1=8

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,8)".And we see that t_{\alpha/2}=

Now we have everything in order to replace into formula (1):

19.1-3.355\frac{1.5}{\sqrt{9}}=17.42    

19.1+3.355\frac{1.5}{\sqrt{9}}=20.78    

And the best option would be:

C. [17.42,20.78]

5 0
3 years ago
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