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According to the U.S. Department of Transportation’s Air Travel Consumer Report, the nation’s 12 largest airlines recorded an on-time arrival percentage of 77.4% during 2001. Of interest is to estimate the mean delay time for all flights that did not arrive on time during 2013. A simple random sample of 35 late arriving flights was selected, and the mean delay time of this sample of 35 flights was 14.2 minutes, with a standard deviation (s) of 6.4 minutes. Use this information to calculate and interpret a 95% confidence interval for the mean delay time for all flights that did not arrive on time during 2013.

Respuesta :

Answer:

[tex]14.2-2.03\frac{6.4}{\sqrt{35}}=12.004[/tex]    

[tex]14.2+2.03\frac{6.4}{\sqrt{35}}=16.396[/tex]    

We are 95% confidence that the true mean for the delay time is between (12.004 and 16.396)

Step-by-step explanation:

Information given

[tex]\bar X=14.2[/tex] represent the sample mean for the delay time

[tex]\mu[/tex] population mean

s=6.4 represent the sample standard deviation

n=35 represent the sample size  

Confidence interval

The confidence interval for the true parameter of interest is given by:

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

The degrees of freedom, given by:

[tex]df=n-1=35-1=34[/tex]

The Confidence level is 0.95 or 95%,and the significance is [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and the critical value for this case is [tex]t_{\alpha/2}=2.03[/tex]

And replacing we got:

[tex]14.2-2.03\frac{6.4}{\sqrt{35}}=12.004[/tex]    

[tex]14.2+2.03\frac{6.4}{\sqrt{35}}=16.396[/tex]    

We are 95% confidence that the true mean for the delay time is between (12.004 and 16.396)

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