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kow [346]
3 years ago
7

lake trail is 4 3/5 miles long.outlook trail is 5 5/6 miles long .pinewoods trail is 1 3/10 miles longer than lake trail.

Mathematics
1 answer:
wolverine [178]3 years ago
6 0
What's the question so I can answer it for you?
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We considered the differences between the temperature readings in January 1 of 1968 and 2008 at 51 locations in the continental
mr Goodwill [35]

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1.1-2.02\frac{4.9}{\sqrt{50}}=-0.30    

1.1+2.02\frac{4.9}{\sqrt{50}}=2.50    

So on this case the 90% confidence interval would be given by (-0.30;2.50)  

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=1.1 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=4.9 represent the sample standard deviation

n=51 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=51-1=50

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,50)".And we see that t_{\alpha/2}=2.02

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

1.1-2.02\frac{4.9}{\sqrt{50}}=-0.30    

1.1+2.02\frac{4.9}{\sqrt{50}}=2.50    

So on this case the 90% confidence interval would be given by (-0.30;2.50)    

8 0
3 years ago
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