Respuesta :
Answer:
Null hypothesis:[tex]\mu \geq 250[/tex]
Alternative hypothesis:[tex]\mu < 250[/tex]
[tex]p_v =P(t_{15}<-1.6)=0.065[/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 FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=510[/tex] represent the mean for the sample
[tex]s=50[/tex] represent the standard deviation for the sample
[tex]n=16[/tex] sample size
[tex]\mu_o =530[/tex] represent the value that we want to test
[tex]\alpha[/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)
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to determine if the claim thet the handle on average is 530 bags per hour:
Null hypothesis:[tex]\mu \geq 530[/tex]
Alternative hypothesis:[tex]\mu < 530[/tex]
We don't know the population deviation and the sample size is lees than 30, so for this case 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{\sigma}{\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".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{510-530}{\frac{50}{\sqrt{16}}}=-1.60[/tex]
4) Calculate the critical value
We need to begin calculating the degrees of freedom
[tex]df=n-1=16-1=15[/tex]
The critical value for this case would be :
[tex]P(t_{15}<a)=0.05[/tex]
The value of a that satisfy this on the normal standard distribution is a=-1.75 and would be the critical value on this case zc=-1.75
5) Calculate the P-value
Since is a one-side upper test the p value would be:
[tex]p_v =P(t_{15}<-1.6)=0.065[/tex]
6) 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 FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.
The considered sample is indicates that the manufacturer’s claim is not overstated. The appropriate hypothesis are:
- Null hypothesis: The manufacture's claim isn't overstated, and that:
- [tex]H_0: \mu = 530[/tex]
- Alternative hypothesis: The manufacture's claim is overstated and machine can handle less than the given figure, thus:
- [tex]H_1: \mu < 530[/tex]
How to form the hypotheses?
There are two hypotheses. First one is called null hypothesis and it is chosen such that it predicts nullity or no change in a thing. It is usually the hypothesis against which we do the test. The hypothesis which we put against null hypothesis is alternate hypothesis.
Null hypothesis is the one which researchers try to disprove.
For this case, we want to check if the claim of population mean being 530 bags per hour is true or not.
Forming hypothesis:
Null hypothesis: The manufacture's claim isn't overstated, and that:
[tex]H_0: \mu = 530[/tex]
Alternative hypothesis: The manufacture's claim is overstated and machine can handle less than the given figure, thus:
[tex]H_1: \mu < 530[/tex]
where [tex]\mu[/tex] is real population mean
Since we've got:
- sample size = n = 16 < 30 (thus, t-test for single mean will be used)
- sample mean = [tex]\overline{x} = 510[/tex]
- sample standard deviation = [tex]s = 50[/tex]
The t statistic for single mean is evaluated as:
[tex]t = \dfrac{\overline{x} - \mu}{\sigma/\sqrt{n}}[/tex] (if population standard deviation [tex]\sigma[/tex] is not available, we can use sample standard deviation 's' there).
Thus, we get:
[tex]t = \dfrac{\overline{x} - \mu}{s/\sqrt{n}} = \dfrac{510 - 530}{50/\sqrt{16}} = -1.6[/tex]
We have level of significance α = .05
and degree of freedom = n-1 = 16 - 1 = 15
At this degree of freedom and level of significance, the critical value of t is obtained from the critical value t-table as: [tex]t_{\alpha/2} = 2.131[/tex]
Since we've got [tex]|t| = 1.6 < t_{\alpha/2}[/tex]null hypothesis and thus, accept the fact that the manufacturer's claim is not overstated. (in case if we'd have got [tex]|t| > t_{\alpha/2}[/tex], then, we'd accept alternative hypothesis by rejecting the null hypothesis)
Thus, the considered sample is indicates that the manufacturer’s claim is not overstated. The appropriate hypothesis are:
- Null hypothesis: The manufacture's claim isn't overstated, and that:
- [tex]H_0: \mu = 530[/tex]
- Alternative hypothesis: The manufacture's claim is overstated and machine can handle less than the given figure, thus:
- [tex]H_1: \mu < 530[/tex]
Learn more about t-test for single mean here:
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