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Answer :
Answer:
[tex]t=\frac{136-140}{\frac{11}{\sqrt{24}}}=-1.781[/tex]
[tex]p_v =P(t_{(23)}<-1.781)=0.044[/tex]
If we compare the p value and the significance level assumed [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 mean is less than 140 pounds at 5% of signficance.
Step-by-step explanation:
Data given and notation
[tex]\bar X=136[/tex] represent the sample mean
[tex]s=11[/tex] represent the sample standard deviation
[tex]n=24[/tex] sample size
[tex]\mu_o =140[/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.
We need to conduct a hypothesis in order to check if the mean is less than 140, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 140[/tex]
Alternative hypothesis:[tex]\mu < 140[/tex]
If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic 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{136-140}{\frac{11}{\sqrt{24}}}=-1.781[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=24-1=23[/tex]
Since is a one left tailed test the p value would be:
[tex]p_v =P(t_{(23)}<-1.781)=0.044[/tex]
Conclusion
If we compare the p value and the significance level assumed [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 mean is less than 140 pounds at 5% of signficance.
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