Question

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

Answer 1

To construct the confidence intervals, we use the formula Lower Limit = Sample Mean - (Critical Value) * Standard Error, and Upper Limit = Sample Mean + (Critical Value) * Standard Error. The lower and upper limits for both the 95% and 99% confidence intervals are calculated using the given sample mean, sample standard deviation, and sample size. The claim that the mean amount of garbage per bin is 54.1985 pounds is false based on both the 95% and 99% confidence intervals.

Step-by-step explanation:

To construct a confidence interval for the mean amount of garbage per bin, we can use the formula:

Lower Limit = Sample Mean - (Critical Value) * Standard Error

Upper Limit = Sample Mean + (Critical Value) * Standard Error

Where the critical value is obtained from the t-distribution table based on the desired confidence level and degrees of freedom, and the standard error is calculated as the sample standard deviation divided by the square root of the sample size.

a) For a 95% confidence interval, the critical value is approximately 2.021 (based on 40-1=39 degrees of freedom). The standard error is 3.9/sqrt(40) = 0.617. Therefore, the lower limit of the 95% confidence interval is:

52.8 - 2.021 * 0.617 = 51.087

b) The upper limit of the 95% confidence interval is:

52.8 + 2.021 * 0.617 = 54.513

c) For a 99% confidence interval, the critical value is approximately 2.704 (based on 40-1=39 degrees of freedom). The standard error is the same as in the previous case. Therefore, the lower limit of the 99% confidence interval is:

52.8 - 2.704 * 0.617 = 50.290

d) The upper limit of the 99% confidence interval is:

52.8 + 2.704 * 0.617 = 55.310

e) A claim that the mean amount of garbage per bin is 54.1985 pounds would be considered false, as this value falls outside both the 95% and 99% confidence intervals.

Answer 2

a)
`52.8-2.02(3.9)/(√(40))=51.554`

b)
`52.8+2.02(3.9)/(√(40))=54.046`

c)
`52.8-2.71(3.9)/(√(40))=51.129`

d)
`52.8+2.71(3.9)/(√(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)(s)/(√(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(3.9)/(√(40))=51.554`

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

`52.8+2.02(3.9)/(√(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(3.9)/(√(40))=51.129`

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

`52.8+2.71(3.9)/(√(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.