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Art [367]
4 years ago
5

A sanitation department is interested in estimating the mean amount of garbage per bin for all bins in the city. In a random sam

ple of 40 bins, the sample mean amount was 52.8 pounds and the sample standard deviation was 3.9 pounds. Construct 95% and 99% confidence intervals for the mean amount of garbage per bin for all bins in the city.
a) What is the lower limit of the 95% interval? Give your answer to three decimal places.b) What is the upper limit of the 95% interval? Give your answer to three decimal places.c) What is the lower limit of the 99% interval? Give your answer to three decimal placesd) What is the upper limit of the 99% interval? Give your answer to three decimal places.e) Consider the claim that the mean amount of garbage per bin is 54.1985 pounds. Is the following statement true or false? The decision about the claim would depend on whether we use a 95% or 99% confidence interval: True/False
Mathematics
1 answer:
soldi70 [24.7K]4 years ago
3 0

Answer:

a) 52.8-2.02\frac{3.9}{\sqrt{40}}=51.554    

b) 52.8+2.02\frac{3.9}{\sqrt{40}}=54.046    

c) 52.8-2.71\frac{3.9}{\sqrt{40}}=51.129  

d) 52.8+2.71\frac{3.9}{\sqrt{40}}=54.471  

e) Yes, depends of the confidence level.

At 95 % of confidence the value 54.1985 pounds is not included on the interval. At 5% of significance the statement is FALSE.

At 99 % of confidence the value 54.1985 pounds is included on the interval. So at 1% of significance the statement is TRUE.

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X =52.8 represent the sample mean for the sample

\mu population mean (variable of interest)

s=3.9 represent the sample standard deviation

n=40 represent the sample size  

a) What is the lower limit of the 95% interval? Give your answer to three decimal places

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=40-1=39

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,39)".And we see that t_{\alpha/2}=2.02

Now we have everything in order to replace into formula (1):

52.8-2.02\frac{3.9}{\sqrt{40}}=51.554    

b) What is the upper limit of the 95% interval? Give your answer to three decimal places

52.8+2.02\frac{3.9}{\sqrt{40}}=54.046    

So on this case the 95% confidence interval would be given by (51.554;54.046)    

c) What is the lower limit of the 99% interval? Give your answer to three decimal places

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,39)".And we see that t_{\alpha/2}=2.71

Now we have everything in order to replace into formula (1):

52.8-2.71\frac{3.9}{\sqrt{40}}=51.129  

d) What is the upper limit of the 99% interval? Give your answer to three decimal places

52.8+2.71\frac{3.9}{\sqrt{40}}=54.471  

So on this case the 99% confidence interval would be given by (51.129;54.471)    

e) Consider the claim that the mean amount of garbage per bin is 54.1985 pounds. Is the following statement true or false? The decision about the claim would depend on whether we use a 95% or 99% confidence interval: True/False

Yes, depends of the confidence level.

At 95 % of confidence the value 54.1985 pounds is not included on the interval. At 5% of significance the statement is FALSE.

At 99 % of confidence the value 54.1985 pounds is included on the interval. So at 1% of significance the statement is TRUE.

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3 years ago
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3 years ago
a) Suppose f"(z) exists on an interval I and f(s) has a zero at three distinct points a &lt; b&lt; c on I. Show there is a pbint
Fynjy0 [20]

Answer:

We can find a root in the average point (a+b+c)/3

Step-By-Step Explanation:

We can use the Rolle Theorem. Since f is two times deribable on both [a,b] and [b,c], then there exists points x in (a,b) and y in (b,c) such that f'(x) = f'(y) = 0. Now, again by using Rolle Theorem, since f' is derivable in [x,y] (because it is a closed interval in I), then there exists s in [x,y] such that f''(s) = 0. This proves a.

The cube (x-a)(x-b)(x-c) has 3 roots, a, b and c and it is two times derivable because it is a polynomial. Hence we can use (a) to ensure that there is a root on I. Nevertheless, we can try to find the root manually:

f'(x) = (x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)

f''(x) = (x-c)+(x-b)+(x-c)+(x-a)+(x-b)+(x-a) = 2(x-a)+2(x-b)+2(x-c) = 2( (x-a) + (x-b) + (x-c) ) = 6x - 2 (a+b+c).

We want x such that

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3x - (a+b+c) = 0

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x = (a+b+c)/3

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3 years ago
PLEASE ANSWER ASAP!!
ExtremeBDS [4]

Answer:

<u>Fred paid US$ 310 to Social Security as a payroll tax</u>

<u>Fred paid US$ 72.50 to Medicare as a payroll tax</u>

<u>Fred paid 18% of his income to the Federal government as an income tax</u>

<u>Fred paid 5% of his income to the state government as an income tax</u>

Step-by-step explanation:

Fred's earnings in June were $ 5,000

A. Payment to Social Security

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B. Payment to Medicare

5,000 * 1.45% = 5,000 * 0.0145 = 72.50

C. Federal Income Tax

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D. State Income Tax

250/5,000 = 0.05 = 5%

3 0
3 years ago
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