Respuesta :
Answer:
[tex]z=\frac{0.467 -0.9}{\sqrt{\frac{0.9(1-0.9)}{15}}}=-5.59[/tex]
[tex]p_v =P(z<-5.59)=1.13x10^{-8}[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] 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
[tex]\hat p=\frac{7}{15}=0.467[/tex] estimated proportion of seeds germinated
[tex]p_o=0.9[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] 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:[tex]p\geq 0.9[/tex]
Alternative hypothesis:[tex]p < 0.9[/tex]
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.467 -0.9}{\sqrt{\frac{0.9(1-0.9)}{15}}}=-5.59[/tex]
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 [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
[tex]p_v =P(z<-5.59)=1.13x10^{-8}[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] 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%
It can be said that at 5% of significance the proportion of germinated seeds is significantly lower than 0.9 or 90%.
Given :
- A seed company claims that the germination rate for their seeds is 90 percent.
- Random sample space of seeds (n) = 15
- Number of seeds germinated (X) = 7
The germination rate of seeds:
([tex]\hat{p}[/tex]) = [tex]\frac{7}{15}=0.467[/tex]
To test the claim that the true proportion of germinated seeds is less than 0.9 or 90%:
Null hypothesis: [tex]p\geq 0.9[/tex]
Alternative hypothesis: [tex]p<0.9[/tex]
To conduct a one-sample proportion z-test,
[tex]z=\frac{\hat{p}-p_{0} }\sqrt{ {\frac{p_{0}(1-p_{0} }{n} } }[/tex]
[tex]z=\sqrt{\frac{0.467-0.9}{\frac{0.9(1-0.9)}{15} } } \\z=-5.59[/tex]
The significance level assumed is [tex]a=5[/tex].
[tex]p_{v} =P(z<-5.59)=3\times 10^{-8}[/tex]
So the p-value obtained was a very low value. Therefore, we can said that at 5% of significance the proportion of germinated seeds is significantly lower than 0.9 or 90%.
For more information, refer to the link given below:
https://brainly.com/question/22867285
