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Answer: f(n) = 0.9 × f(n − 1) + 10, f(0) = 175, n > 0
Step-by-step explanation:
Given: A store had 175 cell phones in the month of January.
Every month, 10% of the cell phones were sold and 10 new cell phones were stocked in the store.
Let n be the number of months
Then when n=0
f(0)=175
After first month , n=1
The number of cell phone in store=
Thus
Similarly we can do till n months, we get
f(n)=0.9×f(n−1)+10, f(0) = 175, n > 0
Answer:
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%