# Econ 112 - test researchers claim - level of significance

A researcher believes that the average size of farms in the U.S. has increased from the 2002 mean of 471 acres.

She took a sample of 23 farms in 2011 to test her belief and found a sample mean of 498.8 acres and a sample standard deviation of 46.9 acres.

At a 5% level of significance, test the researcher's claim. What is your conclusion?

A study by Hewitt Associates showed that 79% of companies offer employees flexible scheduling.

Suppose a researcher believes that in accounting firms this figure is lower.

The researcher randomly selects 415 accounting firms and determines that 310 of these firms have flexible scheduling.

At a 1% level of significance, does your test show enough evidence to conclude that a significantly lower percentage of accounting firms offer employees flexible scheduling?

What is the p-value for this test?

The American Lighting Company developed a new light bulb that it believes will last at least 700 hours on average.

A test is to be conducted using a random sample of 100 bulbs, and a 5% level of significance. Assume that the population standard deviation is 15 hours.

What are the consequences of making a type II error?

What is the probability of making a Type II error if the true population mean is 695 hours?

### Comments for Econ 112 - test researchers claim - level of significance

 Oct 03, 2013 Statistical Research by: Staff Answer Part I A researcher believes that the average size of farms in the U.S. has increased from the 2002 mean of 471 acres. She took a sample of 23 farms in 2011 to test her belief and found a sample mean of 498.8 acres and a sample standard deviation of 46.9 acres. At a 5% level of significance, test the researcher's claim. What is your conclusion? Decide between the following two Hypotheses: H₀ = µ = 471 acres. There is no change in the average number of acres. H₁ = µ > 471 acres. There is an increase in the average number of acres. ---------------------------------------

 Oct 03, 2013 Statistical Research by: Staff --------------------------------------- Part II use a one-tailed test (a) If the z-score is greater than 1.645, the results are significant at the .05 level. H₀ is rejected. (b) If the z-score is less than 1.645, H₀ is accepted. z = (x-bar - µ) / (σ / √(N)) z = (498.8 - 471) / (46.9 / √(23)) z = 2.842731691857 ---------------------------------------

 Oct 03, 2013 Statistical Research by: Staff --------------------------------------- Part III z is greater than 1.645. This makes it highly significant. Conclusion: There is an increase in the average number of acres per farm. A study by Hewitt Associates showed that 79% of companies offer employees flexible scheduling. Suppose a researcher believes that in accounting firms this figure is lower. The researcher randomly selects 415 accounting firms and determines that 310 of these firms have flexible scheduling. At a 1% level of significance, does your test show enough evidence to conclude that a significantly lower percentage of accounting firms offer employees flexible scheduling? What is the p-value for this test? ---------------------------------------

 Oct 03, 2013 Statistical Research by: Staff --------------------------------------- Part IV Decide between the following two Hypotheses: H₀ = p = 79%. There is no difference in the percentage of all companies which offer flexible scheduling and the percentage of accounting firms which offer flexible scheduling. H₁ = p < 79%. The percentage of accounting firms which offer flexible scheduling is less than the percentage of all companies which offer flexible scheduling. paccounting = 310/415 = 74.7% σ = √((up)/N) = √((p)(1-p)/N) σ = √(.747(1-.747)/415) σ = √(.747(.253)/415) = 0.0213 = 2.13% ---------------------------------------

 Oct 03, 2013 Statistical Research by: Staff --------------------------------------- Part V Calculate the z-score (the z-score is the distance, in standard deviations, between the accounting firm sample of 74.7% and the mean value for all firms of 79%) z = (paccounting - pall firms)/ σ pall firms = 79% z = (.747 - .79)/ .0213 = -2.018779342723 P (z < -2.018779342723) = ? looking up a z-table http://www.intmath.com/counting-probability/z-table.php P(Z<−2.02) =0.5−P(0 P(Z<−2.02) =2.17% ---------------------------------------

 Oct 03, 2013 Statistical Research by: Staff --------------------------------------- Part VI At a 1% level of significance, the sample data does not show that a significantly lower percentage of accounting firms offer employees flexible scheduling. The American Lighting Company developed a new light bulb that it believes will last at least 700 hours on average. A test is to be conducted using a random sample of 100 bulbs, and a 5% level of significance. Assume that the population standard deviation is 15 hours. What are the consequences of making a type II error? What is the probability of making a Type II error if the true population mean is 695 hours? ---------------------------------------

 Oct 03, 2013 Statistical Research by: Staff --------------------------------------- Part VII Type II error: stating something is false when it is actually true (for example, tests show that an aircraft engine does not need an overhaul, when it does need an overhaul) A Type II error for this question would mean that test sampling showed that the average mean is less than 700 hours when it is actually equal to 700 hours or more. For a sample of 100 and a 5% level of significance [that is 95% level of confidence], the confidence interval [CI] would be: CI = (mean - (σ/√(n)) * z, mean + (σ/√(n)) * z) σ = 15 hours at a 95% confidence interval for a one-tailed test z = 1.645 X̄ = mean (this value is 695) CI = (X̄ - (σ/√(n)) * z, X̄ + (σ/√(n)) * z) CI = (X̄ - (15/√(100)) * 1.645, X̄ + (15/√(100)) * 1.645) CI = (X̄ - 2.4675, X̄ + 2.4675) ---------------------------------------

 Oct 03, 2013 Statistical Research by: Staff --------------------------------------- Part VIII If the true mean is X̄, the sample results should fall in the interval (X̄ - 2.4675, X̄ + 2.4675) at a 95% confidence interval since X̄ is 695 CI = (695 - 2.4675, 695 + 2.4675) CI = (692.5325, 697.4675) For a sample mean of 700 or greater to be valid, the sample mean must be greater than 697.4675 (which is 1.645 standard deviations away from 695). ---------------------------------------

 Oct 03, 2013 Statistical Research by: Staff --------------------------------------- Part IX What are the consequences of a Type II error? The company could not market the light bulb as a 700 bulb. This may mean shelving the campaign, layoffs, and a reduced competitive advantage. Thanks for writing. Staff www.solving-math-problems.com