Answer :
Answer:
The null hypothesis is [tex]H_0 : p \le 0.50[/tex]
The Alternative hypothesis is [tex]H_a : p > 0.50[/tex]
The test statistics is [tex]Z = 0.985[/tex]
The conclusion
There is no sufficient evidence to draw the conclusion that more than half of those that are used the drug experienced relief
Step-by-step explanation:
From the question we are told that
The number of subjects reported experiencing significant relief from their symptoms is [tex]x = 108[/tex]
The sample size is [tex]n = 202[/tex]
The level of significance is [tex]\alpha = 0.01[/tex]
The objective of this experiment is to know whether more than half of those who took the drug where relieved
The null hypothesis is [tex]H_0 : p \le 0.50[/tex]
The Alternative hypothesis is [tex]H_a : p > 0.50[/tex]
Where p is the population proportion.
Generally the sample proportion. is mathematically represented as
[tex]\r p = \frac{x}{n}[/tex]
[tex]\r p = \frac{108}{202}[/tex]
[tex]\r p = 0.5347[/tex]
Now the test statistics is mathematically represented as
[tex]Z = \frac{\r p - p}{\sqrt{\frac{p(1-p)}{n} }}[/tex]
substituting values
[tex]Z = \frac{0.5347 - 0.50}{\sqrt{\frac{0.50(1-0.50)}{202} }}[/tex]
[tex]Z = 0.985[/tex]
Generally the p-value is mathematically evaluated using Excel formula as
[tex]p-value =1 - (NORMSDIST(Z))[/tex]
[tex]p-value =1 - (NORMSDIST( 0.985))[/tex]
[tex]p-value =1 - 0.8377[/tex]
[tex]p-value =0.162[/tex]
Comparing the p-value to the level of significance we see that the p-value is greater than the level of significance
This means the the null hypothesis would not be rejected
Hence there is no sufficient evidence to draw the conclusion that there are more than half of those that are using the drug experience relief
Testing the hypothesis, it is found that since the p-value of the test is of 0.1611 > 0.01, we do not reject the null hypothesis, hence the conclusion is that there is not enough evidence to conclude that more than half of all those using the drug experience relief.
At the null hypothesis, it is tested if the proportion is of 0.5, that is:
[tex]H_0: p = 0.5[/tex]
At the alternative hypothesis, it is tested if the proportion is of more than 0.5, that is:
[tex]H_1: p > 0.5[/tex]
The condition is that there are more than 10 failures and 10 successes, which is respected, as there are 94 and 108, respectively.
The test statistic is given by:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
In which:
- [tex]\overline{p}[/tex] is the sample proportion.
- p is the proportion tested at the null hypothesis.
- n is the sample size.
For this problem, the parameters are:
[tex]p = 0.5, n = 202, \overline{p} = \frac{108}{202} = 0.5347[/tex]
The value of the test statistic is:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.5347 - 0.5}{\sqrt{\frac{0.5(0.5)}{202}}}[/tex]
[tex]z = 0.99[/tex]
The p-value is the probability of finding a sample proportion above 0.5347, which is 1 subtracted by the p-value of z = 0.99.
- Looking at the z-table, z = 0.99 has a p-value of 0.8389.
1 - 0.8389 = 0.1611.
Since the p-value of the test is of 0.1611 > 0.01, we do not reject the null hypothesis, hence the conclusion is that there is not enough evidence to conclude that more than half of all those using the drug experience relief.
A similar problem is given at https://brainly.com/question/24166849