Question

A college professor claims that the entering class this year appears to be smarter than entering classes from previous years. He tests a random sample of 22 of this year's entering students and finds that their mean 1Q score is 119, with standard deviation of 15. The college records indicate that the mean 1Q score for entering students from previous years is 112. If we assume that the 1Q scores of this year's entering class are normally distributed, is there enough evidence to conclude, at the 0.05 level of significance, that the mean IQ score, , of this year's class is greater than that of previous years? 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. The null hypothesis The alternative hypothesis: D-D OSD D2D The type of test statistic: (Choose one) X The value of the test statistic (Round to at least three decimal places.) The critical value at the 0.05 level of significance: (Round to at least three decimal places.) Can we conclude, using the 0.05 level of significance, that the mean 1Q score of this year's dass is greater than that of previous years? O Yes No

Answer 1

given data are,

sample mean ($\bar{x}$) = 119

sample sd (s) = 15

sample size (n) = 22

the null hypothesis:-

$H_{0}:\mu=112$

the alternative hypothesis:-

$H_{1}:\mu>112$

type of test statistic:-

t test statistic.

[ we are doing t test, because we know the sample sd , not the population sd]

test statistic be:-

$t=\frac{\bar{x}-\mu}{s/\sqrt{n}}=\frac{119-112}{15/\sqrt{22}}=2.189$

degrees of freedom = (n-1) = (22-1) = 21

critical value:-

$t_{critical}=1.721$

[ from t distribution table, for df = 21 ,alpha= 0.05, right tailed test ]

decision:-

$t = 2.189>t_{critical}=1.721$

so, we reject the null hypothesis.

we conclude that,

YES, there is significant evidence to claim that the mean IQ scores of this year's class is greater than that of previous years at 0.05 level of significance.

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