The human resources department of an engineering company gives IQ tests to a randomly selected group of new hires every year. They claimed that the mean IQ score of new hires, μ1 , from this year is greater than or equal to the mean IQ score of new hires, μ2 , from last year. This year, 90 new hires took the test and scored an average of 113.1 points with a standard deviation of 14.3 . Last year, 85 new hires took the IQ test and they scored an average of 116.9 points with a standard deviation of 15.6 . Assume that the population standard deviation of the IQ scores from the current year and the last year can be estimated by the sample standard deviations, since the samples that are used to compute them are quite large. Is there enough evidence to reject the claim of the human resources department, at the 0.1 level of significance? Perform a one-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.) The null hypothesis: H0: The alternative hypothesis: H1: The type of test statistic: (Choose one)ZtChi squareF The value of the test statistic: (Round to at least three decimal places.) The critical value at the 0.1 level of significance: (Round to at least three decimal places.) Can we reject the claim that the mean IQ score of new hires from the current year is greater than or equal to the mean IQ score of new hires from last year? Yes No μ σ p x s p
Let , be the population mean of IQ scores of new hires from current year and be the population mean of IQ scores of new hires from last year
Hypothesis : VS
The test statistic under Ho is ,
Critical Value :
Decision : Here ,
Therefore , fail to reject Ho at 0.01 significance level.
Conclusion : Hence , there is sufficient evidence to conclude that the mean IQ score of new hires from the current year is greater than or equal to the mean IQ score of new hires from last year
Get Answers For Free
Most questions answered within 1 hours.