Question

********I NEED THE BELL SHAPED CURVE, PLEASE DON'T ANSWER IF YOU CAN'T INCLUDE***********

********I NEED THE BELL SHAPED CURVE, PLEASE DON'T ANSWER IF YOU CAN'T INCLUDE***********

********I NEED THE BELL SHAPED CURVE, PLEASE DON'T ANSWER IF YOU CAN'T INCLUDE***********

********I NEED THE BELL SHAPED CURVE, PLEASE DON'T ANSWER IF YOU CAN'T INCLUDE***********

According to the Organization for Economic Co-Operation and Development (OECD), adults in the United States worked an average of 1,805 hours in 2007. Assume the population standard deviation is 395 hours and that a random sample of 70 U.S. adults was selected. a. Calculate the standard error of the mean. b. What is the probability that the sample mean will be more than 1,775 hours? c. What is the probability that the sample mean will be between 1,765 and 1,820 hours? d. Would a sample mean of 1,815 hours support the claim made by the OECD? Explain?

Answer 1

Here population mean, = 1805

population standard deviation, = 395

sample size, n = 70

(a)

Standard error of mean =   = =  47.21153007

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(b)

Let be the sample mean.

Probability of sample mean is more than 1775 = P( > 1775)

Probability that sample mean will be more than 1775 = 0.7389

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(c)

Probability that sample mean is between 1765 and 1820 = P(1765 < < 1820)

Probability that sample mean will be between 1765 and 1820 = 0.4278

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(d)

Given sample mean, = 1815

population standard deviation, = 395

sample size, n = 70

The claim made by the OECD is adults in the United States worked an average of 1,805 hours in 2007.

i.e., population mean, = 1805

We have to test H₀ :   = 1805 against H₁ : 1805

The test statistic is,

  

  

Let = 0.05

Then critical region is |z| > 1.960

Here |z| = 0.212 is less than 1.960

We accept the null hypothesis at 5% level of significance.

A sample mean of 1815 hours will support the claim made by the OECD.

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