Mathematics, 12.12.2019 21:31 papabosue

Aprofessor of women's studies is interested in determining if stress affects the menstrual cycle. ten women are randomly sampled for an experiment and randomly divided into two groups. one of the groups is subjected to high stress for two months while the other lives in a relatively stress-free environment. the professor measures the menstrual cycle (in days) of each woman during the second month. the following data are obtained.

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Answer from: EdgeTheWizard

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

$t=\frac{(20.4 -27.6)-(0)}{\sqrt{\frac{2.074^2}{5}}+\frac{2.702^2}{5}}=-4.73$

df=5+5-2=8

$t_{crit}=\pm 2.306$

Reject H0: stress affects the menstrual cycle

Step-by-step explanation:

The statistic is given by the following formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{s^2_2}{n_2}}$

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 = 0$

Alternative hypothesis: $\mu_1 -\mu_2 \neq 0$

Our notation on this case :

n_1 =5 represent the sample size for group 1

n_2 =2 represent the sample size for group 2

Tha data given is:

Group 1 (High stress) : 20,23,18,19,22

Group 2 (Relatively stress free): 26,31,25,26,30

We can calculate the sample mean and the sample deviation with the following formulas:

$\bar X =\frac{\sum_{i=1}^n x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}$

$\bar X_1 =20.4$ represent the sample mean for the group 1

$\bar X_2 =27.6$ represent the sample mean for the group 2

s_1=2.074 represent the sample standard deviation for group 1

s_2=2.702 represent the sample standard deviation for group 2

And now we can calculate the statistic:

$t=\frac{(20.4 -27.6)-(0)}{\sqrt{\frac{2.074^2}{5}}+\frac{2.702^2}{5}}=-4.73$

Now we can calculate the degrees of freedom given by:

df=5+5-2=8

Now we can calculate the critical value since the confidence is 95% the value for the significance would be $\alpha=1-0.95=0.05$ and the value for $\alpha/2 =0.025$ and if we find a critical value on th t distribution with 8 degrees of freedom that accumulates 0.025 of the area on each tail we got $t_{crit}=\pm 2.306$

Since our calculated value is lower than the critical value, we have enough evidence to reject the null hypothesis. And makes sense say that the difference between the two means are different at 5% of significance.

Reject H0: stress affects the menstrual cycle

Answer from: Quest

the answer is d

step-by-step explanation:

y+3=-4(x-2)

y+3=-4x+8

-3 -3

y=-4x+5

Which of the following equations is equivalent to y+3=-4(x-2)?

Answer from: Quest

describing distributions is one of the key skills you’ll need to earn a high score on the ap statistics exam. if you need proof of this, just flip through some past exam questions, which can be found at the collegeboard website. you’ll notice that the first free response question is almost always a question requiring you to look at a graph and describe it.

step-by-step explanation: just trying to get points sorry : (