Respuesta :
Answer:
[tex]z=\frac{6.4-3.1}{\frac{4.5}{\sqrt{9}}}=2.2[/tex]
[tex]p_v =2*P(Z>2.2)=0.0139[/tex]
If we compare the p value and a 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 lifetime is signficantly different from 3.1.
Step-by-step explanation:
Data given and notation
[tex]\bar X=6.4[/tex] represent the sample mean
[tex]\sigma[/tex] represent the population standard deviation
[tex]n=9[/tex] sample size
[tex]\mu_o =3.1[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
SS=162 represent the sum of squares.
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 apply a two tailed test.
What are H0 and Ha for this study?
Null hypothesis: [tex]\mu = 3.1[/tex]
Alternative hypothesis :[tex]\mu \neq 3.1[/tex]
Compute the test statistic
The statistic for this case is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-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
First we need to calculate the sample deviation given by this formula:
[tex]S= \sqrt{\frac{SS}{n-1}}=\sqrt{\frac{162}{8}}=4.5[/tex]
We can replace in formula (1) the info given like this:
[tex]z=\frac{6.4-3.1}{\frac{4.5}{\sqrt{9}}}=2.2[/tex]
Give the appropriate conclusion for the test
Since is a two tailed test the p value would be:
[tex]p_v =2*P(Z>2.2)=0.0139[/tex]
Conclusion
If we compare the p value and a 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 lifetime is signficantly different from 3.1.
