Answer:
We conclude that the effectiveness is less than the 85% claim the company is making.
Step-by-step explanation:
We are given that a manufacturer of a new medication on the market for Crohn's disease makes a claim that the medication is effective in 85% of people who have the disease.
One hundred seventy-five individuals with Crohn's disease are given the medication, and 135 of them note the medication was effective.
<em />
<em>Let p = </em><u><em>percentage of people who have the Crohn's disease.</em></u>
So, Null Hypothesis,
: p = 85% {means that the effectiveness is equal to the 85% claim the company is making}
Alternate Hypothesis,
: p < 85% {means that the effectiveness is less than the 85% claim the company is making}
The test statistics that would be used here <u>One-sample z proportion</u> <u>statistics</u>;
T.S. =
~ N(0,1)
where,
= sample proportion of individuals who note the medication was effective =
= 0.77
n = sample of individuals with Crohn's disease taken = 175
So, <em><u>test statistics</u></em> = ![\frac{0.77-0.85}{\sqrt{\frac{0.77(1-0.77)}{175} } }](https://tex.z-dn.net/?f=%5Cfrac%7B0.77-0.85%7D%7B%5Csqrt%7B%5Cfrac%7B0.77%281-0.77%29%7D%7B175%7D%20%7D%20%7D)
= -2.515
The value of z test statistics is -2.515.
<em>Now, at 0.05 significance level the z table gives critical value of -1.645 for left-tailed test.</em>
<em>Since our test statistic is less than the critical value of z as -2.515 < -1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>
Therefore, we conclude that the effectiveness is less than the 85% claim the company is making.