7x+9y=-41 and x+3y=1. Solve system of equations

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

Identify the type of communication illustrated in the quotation. "Ladies and gentlemen of the jury, I implore you to weigh the e

Katena32 [7]

Public communication because one person is talking to an audience, and it's not big enough to be mass communication. And it's not group because that consists of three or more communicating a shared goal. Also isn't it because interpersonal is a talking between two people.

6 0

3 years ago

Read 2 more answers

Lucy spent a total of $6.60 on 5 apples and a package of grapes. If the package of grapes cost $2.35, how much did each apple co

erastova [34]

Answer:

Step-by-step explanation:

apple cost = total cost - grapes package cost

5 apples cost = 6.60 - 2.35

= 4.25

per apple cost = 4.25 ÷ 5

= 0.85

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4 0

3 years ago

5 lbs equals how many ounces

fredd [130]

5 pounds = 80 ounces

7 0

3 years ago

I need help please please!!!

wel

Answer:

y = 91

Step-by-step explanation:

y = 5² + 10(5) + 16

y = 25 + 50 + 16

y = 91

4 0

3 years ago