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Alina [70]
2 years ago
6

The germination rate is the rate at which plants begin to grow after the seed is planted.

Mathematics
1 answer:
chubhunter [2.5K]2 years ago
4 0

Answer:

z=\frac{0.467 -0.9}{\sqrt{\frac{0.9(1-0.9)}{15}}}=-5.59  

p_v =P(z  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of germinated seeds is significantly lower than 0.9 or 90%

Step-by-step explanation:

Data given and notation

n=15 represent the random sample taken

X=7 represent the number of seeds germinated

\hat p=\frac{7}{15}=0.467 estimated proportion of seeds germinated

p_o=0.9 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 the true proportion of germinated seeds is less than 0.9 or 90%.:  

Null hypothesis:p\geq 0.9  

Alternative hypothesis:p < 0.9  

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{0.467 -0.9}{\sqrt{\frac{0.9(1-0.9)}{15}}}=-5.59  

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 left tailed test the p value would be:  

p_v =P(z  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of germinated seeds is significantly lower than 0.9 or 90%

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Since the p value i approximately 0 we can conclude that we have enough evidence to say that with this method, the probability of a baby being a boy is significantly greater than 0.5

Step-by-step explanation:

Information given

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Hypothesis to test

The ides is verify if with this method, the probability of a baby being a boy is greater than 0.5, so then the system of hypothesis are.:  

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The statitic is given by:

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Replacing the info we got:

z=\frac{0.815 -0.5}{\sqrt{\frac{0.5(1-0.5)}{211}}}=9.151  

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p_v =P(z>9.151)\approx 0  

Since the p value i approximately 0 we can conclude that we have enough evidence to say that with this method, the probability of a baby being a boy is significantly greater than 0.5

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