Respuesta :
Answer:
[tex]z=\frac{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309[/tex]
[tex]p_v =P(z<-0.309)=0.379[/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 adults with the medication was effective is not significantly less than 0.65
Step-by-step explanation:
Data given and notation
n=180 represent the random sample taken
X=115 represent the adults with the medication was effective
[tex]\hat p=\frac{115}{180}=0.639[/tex] estimated proportion of adults with the medication was effective
[tex]p_o=0.65[/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 true proportion is less than 0.65.:
Null hypothesis:[tex]p \geq 0.65[/tex]
Alternative hypothesis:[tex]p < 0.65[/tex]
When we conduct a proportion test we need to use the z statisitc, 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{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309[/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<-0.309)=0.379[/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 adults with the medication was effective is not significantly less than 0.65
Answer:
H0: p = 0.65
Ha: p < 0.65
Sample proportion = 115 / 180 = 0.6389
Test statistics
z = - p / sqrt( p( 1 -p ) / n)
= 0.6389 - 0.65 / sqrt ( 0.65 * 0.35 / 180)
= -0.31
Critical value at 0.05 level = -1.645
Since test statistics falls in non-rejection region, do not reject H0.
We conclude at 0.05 level that we fail to support the claim.
Step-by-step explanation:
