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Household income The 2015American Community Survey estimated the median

household income for each state. According to ACS, the 90%confidence interval for the 2015median household income in Arizona is \(51,492±\)431.

a. Interpret the confidence interval.

b. Interpret the confidence level.

Short Answer

Expert verified

a. There is 90%assurance that income in Arizona fall between $51061and$51923

b. It is used for the calculation of value of level of significance

Step by step solution

01

Given Information

The confidence interval is given as51,492±431=(51061,51923)for90%.

02

To interpret Confidence Interval

Confidence Interval is calculated as 51,492±431=(51061,51923)

It implies that there is 90%assurance that Arizona's median household income falls between$51061and$51923.

03

Interpreting Confidence Level

  • It gives the probability for values that expects to calculate population parameter.
  • It is used for the calculation of value of level of significanceα.

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Most popular questions from this chapter

One reason for using a t distribution instead of the standard Normal distribution to find critical values when calculating a level C confidence interval for a population mean is that

a. zcan be used only for large samples.

b. zrequires that you know the population standard deviation σ.

c. z requires that you can regard your data as an SRS from the population.

d. z requires that the sample size is less than 10% of the population size.

e. a z critical value will lead to a wider interval than a t critical value.

In a poll conducted by phone,

I. Some people refused to answer questions.

II. People without telephones could not be in the sample.

III. Some people never answered the phone in several calls.

Which of these possible sources of bias is included in the ±2%margin of error announced for the poll?

a. I only

b. II only

c. III only

d. I, II, and III

e. None of these

Got shoes? The class in Exercise 1 wants to estimate the variability in the number of pairs

of shoes that female students have by estimating the population standard deviation σ.

Reporting cheating What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172172undergraduates at a large university: “You witness two students cheating on a quiz. Do you go to the professor?” Only 19answered “Yes.” Assume the conditions for inference are met.

a. Determine the critical valueZ*for a 96%confidence interval for a proportion.

b. Construct a 96%confidence interval for the proportion of all undergraduates at this university who would go to the professor.

c. Interpret the interval from part (b).

Losing weight Refer to Exercise 6.

a. Explain what would happen to the length of the interval if the confidence level was decreased to 90%.

b. How would a 95%confidence interval based on triple the sample size compare to the original 95%interval?

c. As Gallup indicates, the 3percentage point margin of error for this poll includes only sampling variability (what they call “sampling error”). What other potential sources of error (Gallup calls these “non sampling errors”) could affect the accuracy of the 95% confidence interval?

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