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allsm [11]
3 years ago
6

i beg of yall a crown, 5 stars, a heart and a ty comment to answer my questions iv posted on my profile, ill be posting more too

.

Mathematics
1 answer:
Over [174]3 years ago
6 0

Answer:

I would say the answer is $5.00/lb

Have a wonderful day!

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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
saul85 [17]

Solution :

Given :

Sample mean, $\overline X = 34.2$

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 ,

  $\overline X \pm t \times \frac{s}{\sqrt n}$

 $34.2 \pm 2.34 \times \frac{8.2}{\sqrt {129}}$

 (32.52, 35.8)

 95% confidence interval is  (32.52, 35.8)

So with $95 \%$ 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 $95 \%$ 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 $5 \%$ 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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3 years ago
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