Respuesta :
Answer:
Yes. There is enough evidence to support the claim that there was a decrease in voter support for the new candidate after the unfortunate remarks.
Step-by-step explanation:
This is a hypothesis test for a proportion.
The previous proportion, to which we will test the new proportion, was:
[tex]\pi=X_0/n_0=223/415=0.5373[/tex]
The claim is that there was a decrease in voter support for the new candidate after the unfortunate remarks.
Then, the null and alternative hypothesis are:
[tex]H_0: \pi=0.5373\\\\H_a:\pi< 0.5373[/tex]
The significance level is 0.05.
The sample has a size n=630.
The sample proportion of the new sample is p=0.5032.
[tex]p=X/n=317/630=0.5032[/tex]
The standard error of the proportion is:
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.5373*0.4627}{630}}\\\\\\ \sigma_p=\sqrt{0.00039}=0.0199[/tex]
Then, we can calculate the z-statistic as:
[tex]z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.5032-0.5373+0.5/630}{0.0199}=\dfrac{-0.0333}{0.0199}=-1.6766[/tex]
This test is a left-tailed test, so the P-value for this test is calculated as:
[tex]P-value=P(z<-1.6766)=0.0468[/tex]
As the P-value (0.0468) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that there was a decrease in voter support for the new candidate after the unfortunate remarks.
Answer:
[tex]z=\frac{0.503-0.537}{\sqrt{0.517(1-0.517)(\frac{1}{415}+\frac{1}{630})}}=-1.076[/tex]
Since is a left tailed test the p value would be:
[tex]p_v =P(Z<-1.076)=0.141[/tex]
For this case if we use a significance level of 10% we see that [tex]p_v >\alpha[/tex] so then we don;'t have enough evidence to conclude that there was a decrease in voter support for the new candidate after the unfortunate remarks were made
Step-by-step explanation:
Data given and notation
[tex]X_{1}=223[/tex] represent the number of people who support the candidate before
[tex]X_{2}=317[/tex] represent the number of people who support the candidate after
[tex]n_{1}=415[/tex] sample before
[tex]n_{2}=630[/tex] sample after
[tex]p_{1}=\frac{223}{415}=0.537[/tex] represent the proportion before
[tex]p_{2}=\frac{317}{630}=0.503[/tex] represent the proportion after
[tex]\hat p[/tex] represent the pooled estimate of p
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the value for the test (variable of interest)
[tex]\alpha=0.05[/tex] significance level given
Concepts and formulas to use
We need to conduct a hypothesis in order to check if is the proportion after is lower than the proportion before, the system of hypothesis would be:
Null hypothesis:[tex]p_{2} \geq p_{1}[/tex]
Alternative hypothesis:[tex]p_{2} < p_{1}[/tex]
We need to apply a z test to compare proportions, and the statistic is given by:
[tex]z=\frac{p_{2}-p_{1}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex] (1)
Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{223+317}{415+630}=0.517[/tex]
Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.503-0.537}{\sqrt{0.517(1-0.517)(\frac{1}{415}+\frac{1}{630})}}=-1.076[/tex]
Statistical decision
Since is a left tailed test the p value would be:
[tex]p_v =P(Z<-1.076)=0.141[/tex]
For this case if we use a significance level of 10% we see that [tex]p_v >\alpha[/tex] so then we don;'t have enough evidence to conclude that there was a decrease in voter support for the new candidate after the unfortunate remarks were made
