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The mayor is interested in finding a 98% confidence interval for the mean number of pounds of trash per person per week that is generated in the city. The study included 129 residents whose mean number of pounds of trash generated per person per week was 34.2 pounds and the standard deviation was 8.2 pounds.

a. The sampling distribution follows a ______ distribution.
b. With 95% confidence the population mean number of pounds per person per week is between_____ and_____ pounds.
c. If many groups of 120 randomly selected people in the city are studied, then a different confidence interval would be produced from each group. About_____ percent of these confidence intervals will contain the true population mean number of pounds of trash generated per person per week and about________ percent will not contain the true population mean number of pounds of trash generated per person per week.

Sagot :

Solution :

Given :

Sample mean, [tex]$\overline X = 34.2$[/tex]

Sample size, n = 129

Sample standard deviation, s = 8.2

a. Since the population standard deviation is unknown, therefore, we use the t-distribution.

b. Now for 95% confidence level,

   α = 0.05, α/2 = 0.025

  From the t tables, T.INV.2T(α, degree of freedom), we find the t value as

  t =T.INV.2T(0.05, 128) = 2.34

  Taking the positive value of t, we get  

  Confidence interval is ,

  [tex]$\overline X \pm t \times \frac{s}{\sqrt n}$[/tex]

 [tex]$34.2 \pm 2.34 \times \frac{8.2}{\sqrt {129}}$[/tex]

 (32.52, 35.8)

 95% confidence interval is  (32.52, 35.8)

So with [tex]$95 \%$[/tex] confidence of the population of the mean number of the pounds per person per week is between 32.52 pounds and 35.8 pounds.

c. About [tex]$95 \%$[/tex] of confidence intervals which contains the true population of mean number of the pounds of the trash that is generated per person per week and about [tex]$5 \%$[/tex] that doe not contain the true population of mean number of the pounds of trashes generated by per person per week.  

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