Respuesta :
Answer:
[tex]z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9[/tex]
[tex]p_v =P(z>1.9)=0.0287[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.
Step-by-step explanation:
Data given and notation
[tex]\bar X=12.95[/tex] represent the sample mean
[tex]\sigma=3[/tex] represent the population standard deviation for the sample
[tex]n=36[/tex] sample size
[tex]\mu_o =12[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is equal to 12, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 12[/tex]
Alternative hypothesis:[tex]\mu > 12[/tex]
Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9[/tex]
P-value
Since is a right tailed test the p value would be:
[tex]p_v =P(z>1.9)=0.0287[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.
Answer:
Yes, we will reject the claim population mean = 12.
Step-by-step explanation:
It is provided that Standard deviation, [tex]\sigma[/tex] = 3 and sample mean,Xbar = 12.95 .
Let, Null Hypothesis,[tex]H_0[/tex] : mean, [tex]\mu[/tex] = 12
Alternate Hypothesis,[tex]H_1[/tex] : mean, [tex]\mu[/tex] > 12
Since Population is Normal so our Test Statistics will be:
[tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] follows standard normal,N(0,1)
Here, sample size, n = 36.
Test Statistics = [tex]\frac{12.95-12}{\frac{3}{\sqrt{36} } }[/tex] = 1.9
So, at 5% level of significance z % table gives the critical value of 1.6449 and our test statistics is higher than this as 1.6449 < 1.9. So,we have sufficient evidence to reject null hypothesis or accept [tex]H_1[/tex] as our test statistics falls in the rejection region because it is more than 1.6449.
Hence we conclude after testing that we will reject claim of Population mean, μ = 12.
