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The Nielsen Company reported that U.S. residents aged 18 to 24 years spend an average of 32.5 hours per month using the Internet on a computer.13 You wonder if this it true for students at your large university because so many students use their smartphones to access the Internet. You collect an SRS of n=75 students and obtain ¯x=28.5 hours with s=23.1 hours.

Required:
a. Report the 95% confidence interval for μ, the average number of hours per month that students at your university use the Internet on a computer.
b. Use this interval to test whether the average time for students at your university is different from the average reported by Nielsen. Use the 5% significance level. Summarize your results.

Respuesta :

Answer:

a) [tex]28.5-1.993\frac{23.1}{\sqrt{75}}=23.18[/tex]    

[tex]28.5+1.993\frac{23.1}{\sqrt{75}}=33.82[/tex]    

b) For this case since the value 32.5 is in the confidence interval obtained then we can't conclude that the statement by Nielsen is wrong

Step-by-step explanation:

Information given

[tex]\bar X=28.5[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

s=23.1 represent the sample standard deviation

n=75 represent the sample size  

Part a

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)

The degrees of freedom are given by:

[tex]df=n-1=75-1=74[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex] and the critical value would be [tex]t_{\alpha/2}=1.993[/tex]

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

[tex]28.5-1.993\frac{23.1}{\sqrt{75}}=23.18[/tex]    

[tex]28.5+1.993\frac{23.1}{\sqrt{75}}=33.82[/tex]    

Part b

For this case since the value 32.5 is in the confidence interval obtained then we can't conclude that the statement by Nielsen is wrong

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