Respuesta :
Answer:
[tex]z=\frac{0.209 -0.2}{\sqrt{\frac{0.2(1-0.2)}{244}}}=0.351[/tex]
[tex]p_v =P(z>0.351)=0.363[/tex]
So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.01[/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 1% of significance the proportion of subjects with nauseas is not higher than 0.2 or 20% at 1% of significance.
Step-by-step explanation:
Data given and notation
n=244 represent the random sample taken
X=51 represent the subjects with nausea
[tex]\hat p=\frac{51}{244}=0.209[/tex] estimated proportion of subjects with nausea
[tex]p_o=0.2[/tex] is the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level
Confidence=99% or 0.99
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 proportion is higher than 0.2.:
Null hypothesis:[tex]p \leq 0.2[/tex]
Alternative hypothesis:[tex]p > 0.2[/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{0.209 -0.2}{\sqrt{\frac{0.2(1-0.2)}{244}}}=0.351[/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=0.01[/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>0.351)=0.363[/tex]
So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.01[/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 1% of significance the proportion of subjects with nauseas is not higher than 0.2 or 20% at 1% of significance.
Answer:
The null hypothesis, [tex]H_{0}[/tex] should not be rejected
Step-by-step explanation:
Out of 244 subjects, 51 subjects develop nausea:
Sample proportion, [tex]\hat{p} = \frac{51}{244} \\[/tex]
[tex]\hat{p} = 0.209[/tex]
We want to test whether Population proportion, p > 20% or not
i.e whether [tex]p > \frac{20}{100}[/tex] or not
i.e whether p > 0.2 or not
Null hypothesis, H₀ : p ≤ 0.2
Alternative hypothesis, [tex]H_{a} : p > 0.2[/tex]
The test statistic is given by the formula:
[tex]z = \frac{\hat{p} -p_{0} }{\sqrt{\frac{p_{0}(1-p_{0} ) }{n} } }[/tex]
n = 244
[tex]p_{0} = 0.2[/tex]
[tex]1 - p_{0} = 1-0.2 = 0.8[/tex]
[tex]\hat{p}-p_{0} = 0.209 - 0.2 = 0.009[/tex]
[tex]z = \frac{0.009 }{\sqrt{\frac{0.2(0.8 ) }{244} } }[/tex]
z = 0.3515
p(z>0.3515) = 1 - p(z≤0.3515) = 1 - 0.637
p(z>0.3515) = 0.363
p value = 0.363
α = 0.01
Since 0.363 > 0.01
The null hypothesis, [tex]H_{0}[/tex] should not be rejected
There is no evidence that supports the claim that the null hypothesis should be rejected at 0.01 significance level
