Respuesta :
Answer:
[tex]z=\frac{0.54 -0.58}{\sqrt{\frac{0.58(1-0.58)}{820}}}=-2.32[/tex]
[tex]p_v =P(z<-2.32)=0.0102[/tex]
Since the p value is lower than the significance level [tex] \alpha=0.1[/tex] we have enough evidence to reject the null hypothesis and the claim for the manager makes sense.
For the Ti84 preocedure we need to do this:
STAT> TESTS> 1-Z prop Test
And then we need to input the following values:
po= 0.58
x = 443 , n= 820
prop <po
And then calculate and we will get the same results
Step-by-step explanation:
Data given and notation
n=820 represent the random sample taken
X=443 represent the people that had cell phones
[tex]\hat p=\frac{443}{820}=0.540[/tex] estimated proportion of people that had cell phones
[tex]p_o=0.58[/tex] is the value that we want to test
[tex]\alpha=0.1[/tex] represent the significance level
Confidence=90% or 0.90
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
System of hypothesis
We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.58.:
Null hypothesis:[tex]p\geq 0.58[/tex]
Alternative hypothesis:[tex]p < 0.58[/tex]
The statistic is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
And replacing we got:
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.54 -0.58}{\sqrt{\frac{0.58(1-0.58)}{820}}}=-2.32[/tex]
Decision
The significance level provided [tex]\alpha=0.1[/tex]. Now we can calculate the p value
Since is a left tailed test the p value would be:
[tex]p_v =P(z<-2.32)=0.0102[/tex]
Since the p value is lower than the significance level [tex] \alpha=0.1[/tex] we have enough evidence to reject the null hypothesis and the claim for the manager makes sense.
For the Ti84 preocedure we need to do this:
STAT> TESTS> 1-Z prop Test
And then we need to input the following values:
po= 0.58
x = 443 , n= 820
prop <po
And then calculate and we will get the same results
