Respuesta :
Answer:
[tex]z=\frac{0.29 -0.32}{\sqrt{\frac{0.32(1-0.32)}{210}}}=-0.932[/tex]
[tex]p_v =P(z<-0.932)=0.176[/tex]
So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of pet owners contacted by telephone is NOT significantly lower than 0.32
Step-by-step explanation:
Data given and notation
n=210 represent the random sample taken
X=61 represent the number of pet owners contacted by telephone
[tex]\hat p=\frac{61}{210}=0.29[/tex] estimated proportion of pet owners contacted by telephone
[tex]p_o=0.32[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
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 the true proportions is lower than 0.32.:
Null hypothesis:[tex]p \geq 0.32[/tex]
Alternative hypothesis:[tex]p < 0.32[/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{0.29 -0.32}{\sqrt{\frac{0.32(1-0.32)}{210}}}=-0.932[/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 provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
[tex]p_v =P(z<-0.932)=0.176[/tex]
So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of pet owners contacted by telephone is NOT significantly lower than 0.32
