Reflect -4,3 over the x-axis what is it

Erythromycin is a drug that has been proposed to possibly lower the risk of premature delivery. A related area of interest is it

katrin2010 [14]

Answer:

$z=\frac{0.333 -0.3}{\sqrt{\frac{0.3(1-0.3)}{195}}}=1.01$

$p_v =P(z>1.01)=0.156$

So the p value obtained was a very low value and using the significance level asumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of women who complain of nausea between the 24th and 28th week of pregnancy is not significantly higher than 0.3 or 30%

Step-by-step explanation:

Data given and notation

n=195 represent the random sample taken

X=65 represent the women who complain of nausea between the 24th and 28th week of pregnancy

$\hat p=\frac{65}{195}=0.333$ estimated proportion of women who complain of nausea between the 24th and 28th week of pregnancy

$p_o=0.3$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportion is higher than 0.3.:

Null hypothesis: $p\leq 0.3$

Alternative hypothesis: $p > 0.3$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.333 -0.3}{\sqrt{\frac{0.3(1-0.3)}{195}}}=1.01$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level assumed is $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>1.01)=0.156$

So the p value obtained was a very low value and using the significance level asumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of women who complain of nausea between the 24th and 28th week of pregnancy is not significantly higher than 0.3 or 30%

6 0

3 years ago

Which of the following can be used to explain a statement in a geometric proof?

Semenov [28]

Definition, postulate, corollary and theorem

3 0

3 years ago

On the graph of the equation X-4Y= -16 what is the value of the Y-intercept?

Nonamiya [84]

we can always find the x-intercept by simply settting y = 0, and solving for "x".

and we can always find the y-intercept by simply setting x = 0 and solving for "y".

$\bf x-4y=-16\implies \stackrel{x=0}{0-4y=-16}\implies y=\cfrac{-16}{-4}\implies y=4
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8 0

3 years ago

Prove the identity <br> 2sin^2x - 1 = sin^2x - cos^2x

faust18 [17]

Hello,

sin² x-cos² x=sin² x-(1-sin² x)=2sin² x-1
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7 0

3 years ago

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A shipping company charged $15 to ship a package and 5% to insure the package. What was the total cost to ship this particular i

Crank

The shipping would be 15.75 because 5% of 15 is 0.75, so add 15 to 0.75 and the answer is 15.75 :)

5 0

3 years ago

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