Respuesta :
Answer:
A) Yes. At a significance level of 0.05, there is enough evidence to support the claim that the mean water temperature is significantly below 100 °F.
B) P-value = 0.001
C) The probability of not rejecting the null hypothesis at α = 0.05 if the true mean is 104 °F is P(M>98.9)=1. This means that is almost impossible to reject the null hypothesis μ≤100 given that the true mean is 104.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the mean water temperature is significantly below 100 °F.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu=100\\\\H_a:\mu< 100[/tex]
The significance level is 0.05.
The sample has a size n=9.
The sample mean is M=98.
The standard deviation of the population is known and has a value of σ=2.
We can calculate the standard error as:
[tex]\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{2}{\sqrt{9}}=0.667[/tex]
Then, we can calculate the z-statistic as:
[tex]z=\dfrac{M-\mu}{\sigma_M}=\dfrac{98-100}{0.667}=\dfrac{-2}{0.667}=-3[/tex]
This test is a left-tailed test, so the P-value for this test is calculated as:
[tex]\text{P-value}=P(z<-3)=0.001[/tex]
As the P-value (0.001) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that the mean water temperature is significantly below 100 °F.
The critical value for this left-tailed test is zc=-1.645.
The null hypothesis would be accepted if the test statistic is higher than zc=-1.645. For a normal distribution with the parameters of the null hypothesis (μ=100, σ=2), this would correspond to a value of:
[tex]X=\mu+z\sigma/\sqrt{n}=100+(-1.645)\cdot 2/\sqrt{9}=100-1.1=98.9[/tex]
This means that if we get a sample of size n=9 and mean bigger than 98.9, we will failed to reject the null hypothesis.
If the true mean is 104 °F, the probability of getting a sample mean over 98.9 for this sample size can be calculated as:
[tex]z=\dfrac{M-\mu}{\sigma/\sqrt{n}}=\dfrac{98.9-104}{2/\sqrt{9}}=\dfrac{-5.1}{0.6667}=-7.65\\\\\\P(M>98.9)=P(z>-7.65)=1[/tex]
The probability of not rejecting the null hypothesis at α = 0.05 if the true mean is 104 °F is P(M>98.9)=1. This means that is almost impossible to reject the null hypothesis μ≤100 given that the true mean is 104.
