Answer to Question #280381 in Statistics and Probability for gunjan

Question

An educator claims that the average IQ of American college student is at most 110, and

that in a study made to test this claim 150 American college students selected at random had

an average IQ of 111.2 with a standard deviation of 7.2. Use a level of significance of 0.01

to test the claim of the educator.

Answer 1

$n=150,\space s=7.2,\bar{x}=111.2$

The hypotheses tested are,

$H_0:\mu\leq110\space vs\space H_1:\mu\gt110$

The test statistic is given by,

$t={(\bar{x}-\mu)\over ({s\over\sqrt{n}})}={(111.2-110)\over({7.2\over\sqrt{150}})}=2.0412$

$t$ is compared with the t distribution table value at $\alpha=0.01$ with $n-1=150-1=149$ degrees of freedom given as,

$t_{{0.01},149}=t_{0.01,149}= 2.351635$

The null hypothesis is rejected if, $t\gt t_{0.01,149}$

Since $t=2.0412\lt t_{0.01,149}=2.351635,$ we fail to reject the null hypothesis and conclude that there is sufficient evidence to support the educator's claim that the average IQ of American college student is at most 110 at 1% level of significance.