You are studying the water properties of a bog in eastern Wisconsin and wish to estimate its acidity. You de ne a grid that covers the bog and pick 82 points from the grid at random. The mean acidity of your sample is 4.8, and the sd is 1.3. Construct and interpret an 85% con dence interval for the population mean acidity.

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Answer:

[tex]4.8-1.453\frac{1.3}{\sqrt{82}}=4.591[/tex]    

[tex]4.8+1.453\frac{1.3}{\sqrt{82}}=5.009[/tex]    

So on this case the 85% confidence interval would be given by (4.591;5.009)    

And we can conclude that the true mean for the acidity at 85% of confidence is between  (4.591;5.009)    

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[/tex] represent the sample mean for the sample  

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

s represent the sample standard deviation

n 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=82-1=81[/tex]

Since the Confidence is 0.85 or 85%, the value of [tex]\alpha=0.15[/tex] and [tex]\alpha/2 =0.075[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.075,81)".And we see that [tex]t_{\alpha/2}=1.453[/tex]

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

[tex]4.8-1.453\frac{1.3}{\sqrt{82}}=4.591[/tex]    

[tex]4.8+1.453\frac{1.3}{\sqrt{82}}=5.009[/tex]    

So on this case the 85% confidence interval would be given by (4.591;5.009)    

And we can conclude that the true mean for the acidity at 85% of confidence is between  (4.591;5.009)    

Answer:

Confidence Interval (4.59, 5.01)

Step-by-step explanation:

n = 82 (Size)

X = (4.8)

sd = (1.3)

CI = X±Z×sd/√n

So have to look for z value which is given by degrees of freedom and significance level

DF = 82 -1 = 81

SL =100-85 =15%

The Z value is 1.453

CI lower limit = 4.8 - 1.453×1.3/√82 =4.59

    Upper limit = 4.8 + 1.453×1.3/√82 = 5.01

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