Respuesta :
Answer:
I) Null hypothesis:[tex]\mu = 0[/tex]
II) Alternative hypothesis:[tex]\mu <0 [/tex]
III) [tex]t=\frac{-32.5-0}{\frac{466.4}{\sqrt{900}}}=-2.09[/tex]
[tex]p_v =P(Z<-2.09)=0.0183[/tex]
[tex] \alpha=0.05[/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 conclude that the difference is significant less than 0 at 5% of signficance.
[tex]\alpha =0.01[/tex]
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 fail reject the null hypothesis, so we conclude that the difference is not significant less than 0 at 1% of signficance.
IV) A negative difference of -32.5 is significant using confidence levels less than 99% (1% significance) for example at 95% (5% significance) or 97% (3% significance) we have the conclusion of a significant difference, but at 99% of confidence we have the opposite result.
V) Is not appropiate use an statistical test like the t test or the z test in order to determine causality since we have another types of tests in order to see causality. And the causality is not a thing easy to measure because is associated to a lot of factors.
For this case we are assuming to conduct the t test that we use the same individuals in order to get the values before and after and in order that our test would be significant, each observation for the difference follows a normal distribution and the measures are independent from one individual to the other.
Step-by-step explanation:
Data given and notation
[tex]\bar Y=-32.8[/tex] represent the sample mean obtained
[tex]s=466.4[/tex] represent the sample standard deviation for the sample
[tex]n=73[/tex] sample size
[tex]\mu_o =0[/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)
Part I: State the null hypotheses.
We need to conduct a hypothesis in order to check if the null hypothesis is that there was no change in average liquor consumption.
Null hypothesis:[tex]\mu = 0[/tex]
Part II: State the alternative hypothesis
Alternative hypothesis:[tex]\mu <0 [/tex]
Where [te]\mu[/tex] represent the difference between before and after.
If we analyze the size for the sample is > 30 but 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".
Part III: Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{-32.5-0}{\frac{466.4}{\sqrt{900}}}=-2.09[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=900-1=899[/tex]
But isnce the sample size is very large we can assume that the t distribution with 899 degrees of freedom is approximately normal.
Since is a left side test the p value would be:
[tex]p_v =P(Z<-2.09)=0.0183[/tex]
Conclusion
[tex] \alpha=0.05[/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 conclude that the difference is significant less than 0 at 5% of signficance.
[tex]\alpha =0.01[/tex]
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 fail reject the null hypothesis, so we conclude that the difference is not significant less than 0 at 1% of signficance.
Part IV
A negative difference of -32.5 is significant using confidence levels less than 99% (1% significance) for example at 95% (5% significance) or 97% (3% significance) we have the conclusion of a significant difference, but at 99% of confidence we have the opposite result.
Part V
Is not appropiate use an statistical test like the t test or the z test in order to determine causality since we have another types of tests in order to see causality. And the causality is not a thing easy to measure because is associated to a lot of factors.
For this case we are assuming to conduct the t test that we use the same individuals in order to get the values before and after and in order that our test would be significant, each observation for the difference follows a normal distribution and the measures are independent from one individual to the other.
The null hypothesis is that there's no change in liquor consumption.
How to compute the hypothesis?
From the information given, the null hypothesis is that there's no change in liquor consumption. The alternative hypothesis is that the liquor consumption is less than zero.
The t statistic will be:
= ✓900 - (32.8/466.4)
= -2.109
The p value of the test statistic will be:
p = 2 × p(z > t)
p = 2 × p(z > -2.109)
p = 2 × [1 - p(z > -2.109)
p = 2 × (1 - 0.985738)
= 0.028524
The p value of the test statistic shows that at 5% level of significance, the average liquor consumption is declined.
The estimated fall in consumption is large since the mean difference is large by statistical inference.
Lastly, for improved analysis of liquor consumption, it's important to consider the change on the habits of the individual for a time period, reasons for the consumption and its consequences.
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