A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selected nine stu

Question

A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selected nine students. Their mean age was 19.1 years with a sample standard deviation of 1.5 years. What is the 99% confidence interval for the population mean?

A. [0.44,3.80]
B. [14.23,23.98]
C. [17.42,20.78]
D. [17.48,20.72]

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Josie 3 months 2022-02-02T09:42:02+00:00 1 Answer 0 views 0

Answers ( )

    0
    2022-02-02T09:43:41+00:00

    Answer:

    [tex]19.1-3.355\frac{1.5}{\sqrt{9}}=17.42[/tex]    

    [tex]19.1+3.355\frac{1.5}{\sqrt{9}}=20.78[/tex]    

    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”.

    [tex]\bar X=19.1[/tex] represent the sample mean

    [tex]\mu[/tex] 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:

    [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=9-1=8[/tex]

    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: “=-T.INV(0.005,8)”.And we see that [tex]t_{\alpha/2}=[/tex]

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

    [tex]19.1-3.355\frac{1.5}{\sqrt{9}}=17.42[/tex]    

    [tex]19.1+3.355\frac{1.5}{\sqrt{9}}=20.78[/tex]    

    And the best option would be:

    C. [17.42,20.78]

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45:7+7-4:2-5:5*4+35:2 =? ( )