−15x+60≤105 AND14x+11≤−31

In a study, researchers wanted to measure the effect of alcohol on the hippocampal region, the portion of the brain responsibl

schepotkina [342]

Answer:

$t=\frac{8.19-9.02}{\frac{0.8}{\sqrt{17}}}=-4.28$

The degrees of freedom are given by

$df=n-1=17-1=16$

The p value would be given by this probability:

$p_v =P(t_{(16)}$

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

Step-by-step 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:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{8.19-9.02}{\frac{0.8}{\sqrt{17}}}=-4.28$

The degrees of freedom are given by

$df=n-1=17-1=16$

The p value would be given by this probability:

$p_v =P(t_{(16)}$

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

6 0

3 years ago

What fraction of the female children will have hemophilia

Effectus [21]

A woman who is a carrier has: a 25% (one in four) chance of having a daughter who is a carrier. a 25% chance of having a son with hemophilia. and a 50% chance of having a child (boy or girl) who does not have the gene for hemophilia at all

7 0

3 years ago

Solve for n m = -2n-3/kn

Gekata [30.6K]

Answer:

$n=\frac{-km}{2k+3}$