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Anastasy [175]
2 years ago
10

90 gallons increased by 130%

Mathematics
2 answers:
Alex17521 [72]2 years ago
8 0
Multiply 90 by 1.3

The 1.3 is because 130 is bigger than 100 the .3 is for the 30
ASHA 777 [7]2 years ago
5 0
90 * 1.3 = Answer
Answer = 117 gallons
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Erythromycin is a drug that has been proposed to possibly lower the risk of premature delivery. A related area of interest is it
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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%

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