A sample of 41 lots in a suburb has a mean of 192 nesting bi

A sample of 41 lots in a suburb has a mean of 19.2 nesting birds with a standard deviation of 3.5 birds. a. Is s or given?

b. Use your calculator to create a 95% confidence interval estimate for the mean number of nesting birds for lots in this suburb. Record the calculator command used, and write a sentence that interprets your result.

c. With .05 significance, test the claim that the mean number of birds nesting in lots in this suburb is greater than 18. Use your calculator.

d. Create a 99% confidence interval estimate for the standard deviation in number of nesting birds in this suburb. Use formula & chart

Solution

a)

s is given, because this is just a sample. [sigma is for a population]

ANSWER: s

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b)

We use z because n > 30 here.

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    19.2          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    3.5          
n = sample size =    41          
              
Thus,              
Margin of Error E =    1.07133232          
Lower bound =    18.12866768          
Upper bound =    20.27133232          
              
Thus, the confidence interval is              
              
(   18.12866768   ,   20.27133232   ) [ANSWER]

If you are using a ti-84, you may be using ZInterval for this problem.

We are 95% confident that the true mean number of nesting birds per lot is between 18.129 and 20.271.

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c)

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   18  
Ha:    u   >   18  
              
As we can see, this is a    right   tailed test.      
              
Thus, getting the critical z, as alpha =    0.05   ,      
alpha =    0.05          
zcrit =    +   1.644853627      
              
Getting the test statistic, as              
              
X = sample mean =    19.2          
uo = hypothesized mean =    18          
n = sample size =    41          
s = standard deviation =    3.5          
              
Thus, z = (X - uo) * sqrt(n) / s =    2.195356881          
              
Also, the p value is              
              
p =    0.014069004          
              
As P < 0.05, we   REJECT THE NULL HYPOTHESIS.          
There is significant evidence that the true mean number of birds nesting in lots is greater than 18. [ANSWER]

You can use OneSampZTest here.

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d)

As              
              
df = n - 1 =    40          
alpha = (1 - confidence level)/2 =    0.005          
              
Then the critical values for chi^2 are              
              
chi^2(alpha/2) =    66.766          
chi^2(alpha/2) =    20.707          
              
Thus, as              
              
lower bound = (n - 1) s^2 / chi^2(alpha/2) =    7.339064793          
upper bound = (n - 1) s^2 / chi^2(1 - alpha/2) =    23.66349544          
              
Thus, the confidence interval for the variance is              
              
(   7.339064793   ,   23.66349544   )
              
Also, for the standard deviation, getting the square root of the bounds,              
              
(   2.709070836   ,   4.864513895   ) [ANSWER]