题目
Henry's firm is contemplating a switch from spreadsheets to data analytics software primarily because his firm's risk managers believe that the spreadsheets are responsible for too many errors. Certain members of the finance staff function, however, disagree; the hypothesis of these "Excel sympathizers" is that the firm's current spreadsheets contain no more than one bug per sheet. To test their hypothesis, a sample of 36 sheets is carefully analyzed. The sample mean is 1.50 bugs per sheet with a (sample) standard deviation of 0.90. If the desired confidence level is 95.0%, should Henry reject the null hypothesis that the true error rate is not greater than 1.0 bug per sheet?
选项
A.Yes, because the test statistic of 3.33 is greater than the critical value of 1.690
B.Yes, because the test statistic of 10.0 is greater than the critical value of 2.030
C.No, because the test statistic of 10.0 is less than the critical value of 12.706
D.Because the population distribution and variance are both unknown, we do not have a reliable test statistic for the sample mean
答案
A
解析
Yes, because the test statistic of 3.33 is greater than the critical value of 1.690 This is a test of the sample mean where--as is usually the case--we do not know the population's variance. If the sample were small (less than 30), we would not have a reliable test statistic. However, due to the CLT, the sample mean is characterized by the student's t distribution even when the population distribution is non-normal and the population variance is unknown (further, the normal could be used to approximate the student's t). For one-sided 95.0% confidence, the critical t value is 1.690 per the provided lookup table and assuming35 degrees of freedom=TINV (0.950,35) = 1.689572. The standard error is equal to 0.90/√36 = 0.150 and the test statistic is given by(1.50 - 1.0)/SE = (1.50 - 1.0)/0.150 = 0.50/0.150 = 3.33. 这是对样本均值的检验，在通常情况下，我们不知道总体方差。 如果样本很小（少于30个），我们将没有可靠的测试统计数据。但是，由于CLT，即使总体分布非正态且总体方差未知（甚至可以使用正态来近似学生的t），样本均值仍以学生t分布为特征。对于单边95.0％的置信度，每个提供的查找表的临界t值为1.690，并假设35个自由度=TINV（0.950，35）=1.689572。 标准误差等于 0.90/√36 ，测试统计量为（1.50 －1.0）/ SE=（1.50 －1.0）/0.150= 0.50 / 0.150 = 3.33。