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Fofino [41]
4 years ago
15

A simple random survey of residents of a state asks, "Should Mr. Jones be re-elected so that he may continue his policies that h

elp the middle class?"
Which shows how bias in the survey might have affected the results of the survey?
A.
The question in the survey is biased by characterizing how Mr. Jones' policies will help the middle class.

B.
The question in the survey is biased against middle class residents by suggesting that they need help.

C.
The question in the survey is biased in favor of middle class residents by offering them help from Mr. Jones.

D.
Using simple random sampling rather than stratified random sampling will result in an unrepresentative sample.
Mathematics
1 answer:
Andre45 [30]4 years ago
7 0
Answer A seems to make the most sense to me<span />
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Step-by-step explanation:

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A professor of women's studies is interested in determining if stress affects the menstrual cycle. Ten women are randomly sample
BabaBlast [244]

Answer:

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

5 0
3 years ago
which of the following is the result of using the remainder theorem to find F(-1) for the polynomial function F(x) = -x^3+6x^2-4
balandron [24]

Answer:

d). 22

Step-by-step explanation:

add the -1 for x then solve

7 0
3 years ago
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