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A random sample of 44 taxpayers claimed an average of ​$9,700 in medical expenses for the year. Assume the population standard deviation for these deductions was ​$2,379. Construct confidence intervals to estimate the average deduction for the population with the levels of significance shown below. a.5% b.10% c.20% a. The confidence interval with a 5% level of significance has a lower limit of ​$ nothing and an upper limit of ​$ nothing .

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Related textbook solutions

Introductory Statistics
OER Edition, 2013
Openstax
ISBN: 9781938168208

Related questions

. SAT scores are distributed with a mean of 1,500 and a standard deviation of 300. You are interested in estimating the average SAT score of first-year students at your college. If you would like to limit the margin of error of your 95% confidence interval to 25 points, how many students should you sample? Make sure to give a whole number answer

2. We wish to estimate what percent of adult residents in a certain county are parents. Out of 600 adult residents sampled, 354 had kids. Based on this, construct a 95% confidence interval for the proportion, p, of adult residents who are parents in this county.
Give your answers as decimals, to three places.

In mid-July, some businesses like to remind us that the holiday seasons are coming soon. Just like these businesses, the following questions will immerse you in the festive spirit and will (hopefully!) make you forget the heat waves. An article published in 2016 published the results of a survey on Canadians’ behaviour during this period. You are asked to answer the following questions

 

(a) From the data in the article, explain how we can show that the average amount planned to spend in 2016 for the group of 35+ years old is greater than $680,58 in the sample. Assume that there are 450 people that are between 18 and 34 years old in the sample.

(b) Using the article in reference, can we affirm that in 2016 the gap between the average amount planned to spend between the 18-34 years old and the 35+ years old is greater than 275 $ in the population, in favour of the 35+ years old? Assume the sample’s standard deviation is 150 $ among the 18-34 years old and that the sample’s standard deviation is 200 $ among the 35+ years old. You can also assume that the populations are normal. Use a significance level of α = 0, 01.

(c) Build and interpret a 98 % confidence interval that estimates the gap between the average amount planned to spend between the 18-34 years old and the 35+ years old in 2016. Could we use this interval to find the conclusion of the hypothesis test performed in (b)?

Use the following scenario for questions 6 – 8. In 2016-17, the average SAT score of the UVA first year class was 1420 out of 1600. An advisor in the statistics department is interested in determining whether statistics majors from that class have higher SAT scores than the average. He tests H,:µ = 1420 against Hạ: µ > 1420, where u is the mean SAT score for statistics majors from the 2016-17 first year class. In a simple random sample of 30 statistics majors, he finds a sample mean SAT score of 1440 with sample standard deviation of 40.

6. What is the value of the test statistic for this significance test?

a. 0.09

b. 0.5

c. 2.73

d. 20

7. From what distribution is this test statistic?

a. N(0, 1)

b. N(1420, 7.30)

c. t distribution with 29 degrees of freedom

d. t distribution with 28 degrees of freedom

8.) The p-value for this test is 0.005. Which of the following is a proper interpretation of this p- value?

a.There is a 0.5% chance that the null hypothesis is true.

b. If the population mean SAT score really is 1420, there's a 0.5% chance that the sample a. mean SAT score from 30 statistics majors would equal 1440.

c.If the population mean SAT score really is 1420, there's a 0.5% chance that the sample mean SAT score from 30 statistics majors would be 1440 or lower.

d. If the population mean SAT score really is 1420, there's a 0.5% chance that the sample mean SAT score from 30 statistics majors would be 1440 or higher.

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