Question

The family incomes in a town are normally distributed with a mean of $1300 and a standard deviation of $600 per month.

If a given family has a monthly income of $1000, what is the z-score for this family's income? (round to the hundredths place)

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Question 90.6 pts

The family incomes in a town are normally distributed with a mean of $1300 and a standard deviation of $600 per month.

If a given family has a monthly income of $2200, what is the z-score for this family's income? (round to the hundredths place)

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Question 100.6 pts

The family incomes in a town are normally distributed with a mean of $1300 and a standard deviation of $600 per month.

What is the probability that a given family will have a monthly income between $1000 and $2200? Round your answer to the 10 thousandths place.

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Question 111 pts

CONFIDENCE INTERVAL FOR ONE POPULATION PROPORTION WHEN np>5 AND n(p)(1-p)>5

A poll was taken of 1000 employees in Delaware. The employees in the sample were asked if they called in sick at least once/year, when in fact they were not sick. Of the 1000 samples, 200 said that they did.
Find the 98% confidence interval for the proportion of all employees in Delaware who called in sick at least once/year, when in fact they were not sick.

What is the critical value for this confidence interval? Report this value to the hundredths place.  

What is the standard error for this confidence interval? Round to the thousandths place.

Using the rounded standard error, what is the Margin of Error for this confidence interval? Round to the hundredths place.

The general statement of interpretation for this confidence interval is:

We can be 98% confident that the proportion of all Delaware employees who call in sick at least once/year, when in fact they were not sick, lies between LCL and UCL.

Use the rounded margin of error to find LCL and UCL. Report both confidence limits to the hundredths place.

What is the value of LCL?   

What is the value of UCL?   

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Question 121 pts

CONFIDENCE INTERVAL FOR ONE POPULATION MEAN: SIGMA IS UNKNOWN AND SAMPLE SIZE IS LARGE.

A random sample of 33 pickpocket offenses resulted in victims incurring an average loss of $500.00, with a standard deviation of $125.00.
Find the 95% confidence interval for the average loss for all pickpocket victims.

What is the critical value for this confidence interval? Report to the hundredths place.

What is the standard error for this confidence interval? Round to the 10 thousandths place.

Using the rounded standard error, what is the Margin of Error for this confidence interval? Round to the hundredths place.

The general statement of interpretation for this problem is:

We are 95% certain that the average loss for all pickpocket victims lies between $LCL and $UCL.

Use the rounded margin of error to find LCL and UCL. Round the confidence limits to the hundredths place.

What is the value of LCL? $

What is the value of UCL? $   

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Question 131 pts

CONFIDENCE INTERVAL FOR DIFFERENCE IN 2 POPULATION PROPORTIONS (Large Independent Samples)

A random sample of 100 civilian laborers were selected in Finland (sample 1) and a random sample of 75 civilian laborers were selected from Denmark (sample 2). Of those sample in Finland, 10 were unemployed and of those sampled in Denmark, 6 were unemployed.

Find the 95% confidence interval for the difference in unemployment rates between Finland and Denmark.

The estimate of (p₁ - p₂) =  .(do not round).

What is the critical value for this confidence interval? Report this value to the hundredths place.

What is the standard error for this confidence interval?  Round to the 10 thousandths place.

Using the rounded standard error, what is the Margin of Error for this confidence interval?  Round to the hundredths place.

Use the rounded margin of error to find LCL and UCL. Report the confidence limits to the hundredths place.

What is the value of LCL?   

What is the value of UCL?   

The interpretation of this interval involves 3 steps:

Make a statement of confidence.

Evaluate if the interval is positive, negative, or if it contains zero.

Make a final conclusion about p1 and p2.

We are 95% confident that the difference in the unemployment rates of Finland and Denmark lies between LCL and UCL.

For the following statements use one of the following terms to fill in the blank: greater than, less than, similar.

The confidence interval contains zero, which implies that p1 is   to p2.

Therefore, one can be 95% confident that the unemployment rate in Finland is  to the unemployment

                   rate in Denmark.

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Question 141 pts

CONFIDENCE INTERVAL FOR THE DIFFERENCE IN TWO POPULATION MEANS (Large Independent Samples)

The AAUP conducted a study on faculty salaries between private universities (population 1) and public universities (population 2). In a random sample of 35 private universities (sample 1), the average salary was $88,190.00, with a standard deviation of $28,170.00. In a random sample of 31 public universities (sample 2) the average salary was $73,180.00, with a standard deviation of $26,044.00.

Find the 90% confidence interval for the difference in the average salaries of faculty in private and public universities.

The estimate of (µ₁ - µ₂) =  (do not round)

The critical value is  . Report to the thousandths place.

The standard error is $  . Round to the hundredths place.

Using the rounded standard error, the calculated margin of error is $  . Round to the hundredths place.

Use the rounded margin of error to find LCL and UCL. Round the confidence limits to the hundredths place.

What is the value of LCL? $   

What is the value of UCL? $   

The interpretation of this interval involves 3 steps:

Make a statement of confidence.

Evaluate if the interval is positive, negative, or if it contains zero.

Make a final conclusion about µ₁ and µ₂.

I am 90% confident that the difference in the average faculty salaries in private and public universities lies between $LCL and $UCL.

Use one of the following terms to fill in the blank in the following statements: larger than, smaller than, similar.

The confidence interval is entirely positive, which implies that µ₁ is   µ₂.

Therefore, one can be 90% confident that the average faculty salaries in private universities are  the average faculty salaries in public universities.

Answer 1

Solution:

Given that mean of $1300 .

Standard deviation of $600

given family has a monthly income of $1000,

z = 1000 - 1300 / 600 = -0.5

P( z = -0.5 ) = 0.3085

Question 90

Given that mean of $1300 .

Standard deviation of $600

given family has a monthly income of $2200,

z = 2200 - 1300 / 600 = -1.5

P( z = -1.5 ) = 0.0668

Question 100

Given that mean of $1300 .

Standard deviation of $600

from the above we get

P ( 1000<X<2200 )=P ( −0.5<Z<1.5 )

standard normal table to conclude that:

P ( −0.5<Z<1.5 )=0.6247

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