Respuesta :
Answer:
[tex]p_v =P(z>2.5990)=0.0047[/tex]
So the p value obtained was a very low value and using the significance level asumed [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who have smartphone is ignificantly higher than 0.33
Step-by-step explanation:
Assuming that the question is : "Conduct and test the hypothesis required" we have this
Data given and notation
n=349 represent the random sample taken
X=138 represent the adults who have a smart phone
[tex]\hat p=\frac{138}{349}=0.395415[/tex] estimated proportion of adults who have smartphone
[tex]p_o=0.33[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that 70% of adults say that it is morally wrong to not report all income on tax returns.:
Null hypothesis:[tex]p\leq 0.33[/tex]
Alternative hypothesis:[tex]p > 0.33[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.395415 -0.33}{\sqrt{\frac{0.33(1-0.33)}{349}}}=2.5990[/tex]
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level asumed for this case is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
[tex]p_v =P(z>2.5990)=0.0047[/tex]
So the p value obtained was a very low value and using the significance level asumed [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who have smartphone is ignificantly higher than 0.33
Reject the null hypothesis, and it can be said that at 5% of significance the proportion of adults who have smartphones is significantly higher than 0.33.
Given :
- According to a Pew Research Center study, in May 2011, 33% of all American adults had a smartphone.
- She selects 349 community college students at random and finds that 138 of them have a smartphone. In testing the hypotheses: [tex]\rm H_0: p = 0.33[/tex] verses [tex]\rm H_a: p>0.33[/tex].
The hypothesis is given by:
Null Hypothesis - [tex]\rm H_0: p = 0.33[/tex]
Alternate Hypothesis - [tex]\rm H_a: p>0.33[/tex]
Now, the formula of z statistics is given by:
[tex]z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}} }[/tex]
[tex]z = \dfrac{0.395415-0.33}{\sqrt{\dfrac{0.33(1-0.33)}{349}} }[/tex]
Simplify the above expression.
z = 2.5990
So, the p-value is given by:
[tex]\rm p_v = P(z>2.5990)=0.0047[/tex]
So, it can be concluded from the above calculation to reject the null hypothesis, and it can be said that at 5% of significance the proportion of adults who have smartphones is significantly higher than 0.33.
For more information, refer to the link given below:
https://brainly.com/question/22826675
