Question

In a​ study, researchers wanted to measure the effect of alcohol on the hippocampal​ region, the portion of the brain responsible for​ long-term memory​ storage, in adolescents. The researchers randomly selected 17 adolescents with alcohol use disorders to determine whether the hippocampal volumes in the alcoholic adolescents were less than the normal volume of 9.02 cm cubed. An analysis of the sample data revealed that the hippocampal volume is approximately normal with no outliers and x overbarequals8.19 cm cubed and sequals0.8 cm cubed. Conduct the appropriate test at the alphaequals0.01 level of significance.

Answer 1

`t=(8.19-9.02)/((0.8)/(√(17)))=-4.28`

The degrees of freedom are given by

`df=n-1=17-1=16`

The p value would be given by this probability:

`p_v =P(t_((16))<-4.28)=0.000287`

Since the p value is lower than the significance level provided of 0.01 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 9.02 cm^3

Explanation:

Information given

`\bar X=8.19 cm^3` represent the sample mean

`s=0.8 cm^3` represent the sample deviation

`n=17` sample size

`\mu_o =9.02` represent the value to verify

`\alpha=0.01` represent the significance level for the hypothesis test.

t would represent the statistic

`p_v` represent the p value

System of hypothesis to check

We want to check if the true mean is less than the normal value of 9.02 cm^3, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 9.02`

Alternative hypothesis:
`\mu < 9.02`

The statistic for this case is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the info given we got:

`t=(8.19-9.02)/((0.8)/(√(17)))=-4.28`

The degrees of freedom are given by

`df=n-1=17-1=16`

The p value would be given by this probability:

`p_v =P(t_((16))<-4.28)=0.000287`

Since the p value is lower than the significance level provided of 0.01 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 9.02 cm^3