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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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