风子sophia

真的是谢了

IHideMyself

看看，谢谢lz

szszszszmwt

好帖子，谢谢楼主

neoorangelj

谢谢～～～～

thomas7788

Q1. In order to test if the mean IQ of employees in an organization is greater than 100, a sample of 30 employees is taken. The sample value of the computed z-statistic = 3.4. The appropriate decision at a 5% significance level is to:
A)   reject the null hypothesis and conclude that the population mean is not equal to 100.
B)   reject the null hypotheses and conclude that the population mean is greater than 100.
C)   reject the null hypothesis and conclude that the population mean is equal to 100.
Correct answer is B)
Ho:µ ≤ 100; Ha: µ > 100. Reject the null since z = 3.4 > 1.65 (critical value).
Q2. Maria Huffman is the Vice President of Human Resources for a large regional car rental company. Last year, she hired Graham Brickley as Manager of Employee Retention. Part of the compensation package was the chance to earn one of the following two bonuses: if Brickley can reduce turnover to less than 30%, he will receive a 25% bonus. If he can reduce turnover to less than 25%, he will receive a 50% bonus (using a significance level of 10%). The population of turnover rates is normally distributed. The population standard deviation of turnover rates is 1.5%. A recent sample of 100 branch offices resulted in an average turnover rate of 24.2%. Which of the following statements is most accurate?
A)   For the 50% bonus level, the test statistic is -5.33 and Huffman should give Brickley a 50% bonus.
B)   For the 50% bonus level, the critical value is -1.65 and Huffman should give Brickley a 50% bonus.
C)   Brickley should not receive either bonus.
Correct answer is A)
Using the process of Hypothesis testing:
Step 1: State the Hypothesis. For 25% bonus level - Ho: m ≥ 30% Ha: m < 30%; For 50% bonus level - Ho: m ≥ 25% Ha: m < 25%.
Step 2: Select Appropriate Test Statistic. Here, we have a normally distributed population with a known variance (standard deviation is the square root of the variance) and a large sample size (greater than 30.) Thus, we will use the z-statistic.
Step 3: Specify the Level of Significance. α = 0.10.
Step 4: State the Decision Rule. This is a one-tailed test. The critical value for this question will be the z-statistic that corresponds to an α of 0.10, or an area to the left of the mean of 40% (with 50% to the right of the mean). Using the z-table (normal table), we determine that the appropriate critical value = -1.28 (Remember that we highly recommend that you have the “common” z-statistics memorized!) Thus, we will reject the null hypothesis if the calculated test statistic is less than -1.28.
Step 5: Calculate sample (test) statistics. Z (for 50% bonus) = (24.2 – 25) / (1.5 / √ 100) = −5.333. Z (for 25% bonus) = (24.2 – 30) / (1.5 / √ 100) = −38.67.
Step 6: Make a decision. Reject the null hypothesis for both the 25% and 50% bonus level because the test statistic is less than the critical value. Thus, Huffman should give Soberg a 50% bonus.
The other statements are false. The critical value of –1.28 is based on the significance level, and is thus the same for both the 50% and 25% bonus levels.
Q3. Which of the following statements about test statistics is least accurate?
A)   In the case of a test of the difference in means of two independent samples, we use a t-distributed test statistic.
B)   In a test of the population mean, if the population variance is unknown and the sample is small, we should use a z-distributed test statistic.
C)   In a test of the population mean, if the population variance is unknown, we should use a t-distributed test statistic.
Correct answer is B)         
If the population sampled has a known variance, the z-test is the correct test to use. In general, a t-test is used to test the mean of a population when the population is unknown. Note that in special cases when the sample is extremely large, the z-test may be used in place of the t-test, but the t-test is considered to be the test of choice when the population variance is unknown. A t-test is also used to test the difference between two population means while an F-test is used to compare differences between the variances of two populations.
Q4. In a test of the mean of a population, if the population variance is:
A)     known, a t-distributed test statistic is appropriate.
B)     known, a z-distributed test statistic is appropriate.
C)     unknown, a z-distributed test statistic is appropriate.
Correct answer is B)         
If the population sampled has a known variance, the z-test is the correct test to use. In general, a t-test is used to test the mean of a population when the population variance is unknown. Note that in special cases when the sample is extremely large, the z-test may be used in place of the t-test, but the t-test is considered to be the test of choice when the population variance is unknown

tlight90

thank fo your sharing.

chika

thanks for sharing~

马杀鸡会疯

COOL!THANKS!

amucker

樓主，太感謝你了！

27heaton

谢谢搂住