Answer :
Answer: We reject the null hypothesis.
Step-by-step explanation:
Since we have given that
n = 223
Number of subjects developed nausea = 50
[tex]\hat{p}=\dfrac{50}{238}=0.21[/tex]
We need to claim that more than 20% of users develop nausea.
So, Hypothesis:
[tex]H_0:p\leq 0.20\\\\H_a:p<0.20[/tex]
test statistics would be
[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.21-0.20}{\sqrt{\dfrac{0.2\times 0.8}{223}}}\\\\z=\dfrac{0.01}{0.0268}\\\\z=0.373[/tex]
At 0.10 level of significance, z = 1.28
Since 0.373<1.28 so, test statistic is less than critical value.
Hence, we reject the null hypothesis.
Answer:
Test statistics is less than critical value.
Therefore, we reject the null hypothesis.
Step-by-step explanation:
Given :
Total number of subjects be n,
n =223
Number of subjects that developed nausea = 50
Calculation :
[tex]\hat{p} = \dfrac {50}{223} = 0.22[/tex]
We have to claim that more than 20% of users develop nausea.
[tex]\rm H_0 : p \leq 0.20[/tex]
[tex]\rm H_a : p<0.20[/tex]
Now test statistics,
[tex]z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}} }[/tex]
[tex]z= \dfrac {0.22-0.20}{\sqrt{\dfrac{0.2\times0.8}{223}} }[/tex]
[tex]z=0.746[/tex]
Now, at 0.10 level of significance, z = 1.28
Since, 0.746 < 1.28
So, test statistics is less than critical value.
Therefore, we reject the null hypothesis.
For more information, refer the link given below
https://brainly.com/question/17173491?referrer=searchResults