Question: (30 points) *I just need part "f" of this problem.*

Consider the following sample of annual income (in thousands of dollars) of 10 randomly selected residents of Sweetwater County:

36      45      17      25      63      88      117    74      48      40

1. Calculate the sample mean and median. Interpret the median.
   Sample Mean:
   17+25+36+40+45+48+63+74+88+117= 553/10=55.3
   The mean annual income is $55,300
   Sample Median:
   17,25,36,40, 45,48 ,63,74,88,117
   45 + 48= 93 /2= 46.5
   The median annual income of this sample is $46,500. The number has not been influenced by an outliers in the data set and is the central tendency for this sample of data.
2. Calculate the sample standard deviation (rounded to 2 decimal places), and interpret the result.
   
   x x²
   17 289
   25 625
   36 1296
   40 1600
   45 2025
   48 2304
   63 3969
   74 5476
   88 7744
   117 13689
   553 39,017
   39,017-(553)²/10=8436.1/9=937.34
   √937.34= 30.62
   Thus, the standard deviation for the 10 Sweetwater County residents in the sample is $30,620.
3. Locate the first and third quartiles, and interpret the results.
   Q1= 25+36=61/2=30.5
   Q2= median= 46.5
   Q3=74+88=162/2= 81
   The value of Q1= $30,500 indicated that 25% of the residents in the sample had incomes less than $30,500 and that 75% of them had incomes higher than $30,500. Half of the residents in the same had incomes below $46,500 and the other half had incomes above that amount. The value of Q3= $81,000 indicated that 75% of the residents had incomes less than $81,000 and 25% of the residents in the sample had incomes higher than $81,000.
4. Construct a box-plot of the sample data. Are there any outliers? Why or why not?
   
   IQR= Q3-Q1= 81-30.5=50.5
   3x50.5=151.5
   Lower inner fence= 30.5- 151.5= -121
   Upper inner fence= 151.5 + 81=232.5
   Since none of the data fall below -121 or above 232.5 then there are no outliers.
5. Determine a 95% confidence interval for the corresponding population mean, and interpret the result.
   zvalue= 1.96
   standard deviation of the mean= σ/√n= 30.62/√10=9.68
   mean=55.3
   mean+/- zσ= 55.3+/- 1.96 (9.68)= 55.3+/- 18.9728
   55.3-18.9728=36.33
   55.3+18.9728= 74.27
   Thus I am 95% confident that the population mean annual income is between $36,330 and $74,270.
6. Is there evidence that the population mean annual income is below $60,000? To answer this question, state the hypotheses, determine the rejection and non-rejection regions, calculate the appropriate test statistic, estimate the p-value, and interpret the results.

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