Question

A psychologist wanted to test the belief that eating fish makes one smarter. she took a random sample of 12 persons took them through a fish oil supplement for one year and then assessed their IQ. her results were as follows 116,111,101,120,99,94,106,115,107,101,110,92. test at 5% the hypothesis that the fish oil supplement made a significant increase in the average IQ. The average IQ is expected to be 100.

Answer 1

$$mean =\bar{x}=\dfrac{1}{12}(116+111+101+120+99$$

$$+94+106+115+107+101+110+92)=106$$

$$Variance=s^2=\dfrac{1}{12-1}((116-106)^2$$

$$+(111-106)^2+(101-106)^2+(120-106)^2$$

$$+(99-106)^2+(94-106)^2+(106-106)^2$$

$$+(115-106)^2+(107-106)^2+(101-106)^2$$

$$+(110-106)^2+(92-106)^2)=78$$

$$s=\sqrt{s^2}=\sqrt{78}\approx8.83176$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu\leq100$

$H_1:\mu>100$

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha=0.05,$

$df=n-1=12-1=11$ degrees of freedom, and the critical value for a right-tailed test is $t_c=1.795885.$

The rejection region for this right-tailed test is$R=\{t:t>1.795885\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{106-100}{8.83176/\sqrt{11}}$$

$$\approx2.353394$$

Since it is observed that $t=2.353394>1.795885=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for $\alpha=0.05, df=11,$ $t=2.353394,$ right-tailed is $p=0.019129,$ and since $p=0.019129<0.05=\alpha,$ t is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is greater than $100,$ at the $\alpha=0.05$ significance level.