The germination rate is the rate at which plants begin to grow after the seed is planted.
A seed company claims that the germination rate for their seeds is 90 percent. Concerned that the germination rate is actually less than 90 percent, a botanist obtained a random sample of 15 seeds, of which only 7 seeds germinated.
(a) Is there convincing evidence that the germination rate is less than 90%?

So the p value obtained was a very low value and using the significance level given we have 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

estimated proportion of seeds germinated

is the value that we want to test

represent the significance level

z would represent the statistic (variable of interest)

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:  

Alternative hypothesis:  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

(1)  

The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value .

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

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 . The next step would be calculate the p value for this test.  

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

So the p value obtained was a very low value and using the significance level given we have 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%

your question is incomplete, write all the numbers in the sequence that it tells you.  

step-by-step explanation: