Answer:
There is not evidence that the population mean waiting time is different from 3.9 minutes.
Step-by-step explanation:
From the question we know that the size (n), the mean (x) and the standard deviation (s) of the sample are 81, 4.05 minutes and 0.9 minutes. Additionally, we are going to decide if the waiting time is different or not form 3.9 minutes, so the null and alternative hypotheses are:
H0: m=3.9
H1: m≠3.9
Where m is the mean of the population.
Then, <em>we don't need to be concerned about the shape of the population distribution because the value of the n is bigger than 30</em> and we can use the statistic z as:
So, replacing the values, the test statistic is:
On the other hand, the p value for this test is calculated as:
p value = 2P(z>1.50) = 2(0.0668) = 0.134
Taking into account that the p value is bigger than the level of significance 0.01, the null hypothesis is not reject, and there is not evidence that the population mean waiting time is different from 3.9 minutes.
Answer:
Any way the monkey eats minimum 12 banas on each day of 8 days amointing=12*8=96 bananas
Excess bananas eaten=112–96=16
Monkey eats 8 bananas more on sunny days
number of sunny days=16/8=2 days
Number of sunny days in these 8 days=2 days
500=350(1+0.04/12)^12x
Solve for x
500/350=(1+0.04/12)^12x
X=(Log(500/350)/log(1+0.04/12))/12
X=9 months
Answer: 15.67
Step-by-step explanation:
Pot Santa