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Nataly_w [17]
3 years ago
7

Richard can mow the lawn three times as fast as Terri. Together, they finish mowing the lawn in 20 minutes. Which of the followi

ng equations would describe how many minutes it would take Richard to mow the lawn by himself? a. 1/x+1/3x=1/20 b. 1/x+1/3=1/20 c. 20 (1/x+1/3x)=1/20 d. x+3x=1
Mathematics
1 answer:
marysya [2.9K]3 years ago
7 0

Answer:

d it adds the two up and only mutiplys terris amount.

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A manufacturer of a new medication on the market for Alzheimer's disease makes a claim that the medication is effective in 65% o
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Answer:

z=\frac{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309  

p_v =P(z  

So the p value obtained was a very high value and using the significance level given \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 adults with the medication was effective is not significantly less than 0.65

Step-by-step explanation:

Data given and notation

n=180 represent the random sample taken

X=115 represent the adults with the medication was effective

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p_o=0.65 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

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 less than 0.65.:  

Null hypothesis:p \geq 0.65  

Alternative hypothesis:p < 0.65  

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{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309  

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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 provided \alpha=0.05. The next step would be calculate the p value for this test.  

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

p_v =P(z  

So the p value obtained was a very high value and using the significance level given \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 adults with the medication was effective is not significantly less than 0.65

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