Respuesta :
Answer:
[tex]z=\frac{0.231 -0.2}{\sqrt{\frac{0.2(1-0.2)}{221}}}=1.152[/tex]
The p avlue for this case is given by:
[tex]p_v =P(z>1.152)=0.125[/tex]
The p value for this case is higher than the significance level so then we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true proportion is higher than 0.2 or 20%
Step-by-step explanation:
Information provided
n=221 represent the random sample taken
X=51 represent the people with nausea
[tex]\hat p=\frac{51}{221}=0.231[/tex] estimated proportion of people with nausea
[tex]p_o=0.21[/tex] is the value to test
[tex]\alpha=0.1[/tex] represent the significance level
z would represent the statistic
[tex]p_v[/tex] represent the p value
Hypothesis to verify
We want to check if the true population is higher than 0.20, the system of hypothesis are.:
Null hypothesis:[tex]p \leq 0.2[/tex]
Alternative hypothesis:[tex]p > 0.2[/tex]
The statistic is given:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
Replacing the info given we got:
[tex]z=\frac{0.231 -0.2}{\sqrt{\frac{0.2(1-0.2)}{221}}}=1.152[/tex]
The p avlue for this case is given by:
[tex]p_v =P(z>1.152)=0.125[/tex]
The p value for this case is higher than the significance level so then we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true proportion is higher than 0.2 or 20%
