Respuesta :
Answer:
Null hypothesis: [tex]\mu \geq 10[/tex]
Alternative hypothesis :[tex]\mu<10[/tex]
[tex]t=\frac{9.11-10}{\frac{7.30}{\sqrt{2000}}}=-5.45[/tex]
[tex]p_v =P(t_{(1999)}<-5.45)=2.830x10^{-8}[/tex]
And we can use the following excel code: "=T.DIST(-5.45,1999,TRUE)"
Step-by-step explanation:
Data given and notation
[tex]\bar X=9.11[/tex] represent the sample mean
[tex]s=7.30[/tex] represent the sample standard deviation
[tex]n=2000[/tex] sample size
[tex]\mu_o =5.70[/tex] represent the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.
t 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.
Is a one tailed left test.
What are H0 and Ha for this study?
Null hypothesis: [tex]\mu \geq 10[/tex]
Alternative hypothesis :[tex]\mu<10[/tex]
Find the rejection region for the test using significance level=.01
First we need to find the degrees of freedom on this case:
[tex]df=n-1=2000-1=1999[/tex]
In order to find the critical value we need to find a value on the t distirbution such that:
[tex]P(t_{(1999)}<a)=0.01[/tex] where 0.01 represent the significance level selected.
And this value is [tex]t_{crit}=-2.33[/tex]
We can use the following excel code: "=T.INV(0.01,1999)"
The rejection regin would be (-infinity;-2.33)
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
We can replace in formula (1) the info given like this:
[tex]t=\frac{9.11-10}{\frac{7.30}{\sqrt{2000}}}=-5.45[/tex]
Give the appropriate conclusion for the test
Since is a one side left tailed test the p value would be:
[tex]p_v =P(t_{(1999)}<-5.45)=2.830x10^{-8}[/tex]
And we can use the following excel code: "=T.DIST(-5.45,1999,TRUE)"
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.
R code
