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in a clinical trial, 16 out of 872 patients taking a prescription drug …

Question

in a clinical trial, 16 out of 872 patients taking a prescription drug daily complained of flulike symptoms. suppose that it is known that 1.5% of patients taking competing drugs complain of flulike symptoms. is there sufficient evidence to conclude that more than 1.5% of this drugs users experience flulike symptoms as a side effect at the α = 0.1 level of significance?
(type integers or decimals. do not round.)
find the test statistic, ( z_0 )
( z_0 = 0.81 ) (round to two decimal places as needed)
find the p - value
p - value = 0.209 (round to three decimal places as needed)
choose the correct conclusion below
a. since p - value < α, reject the null hypothesis and conclude that there is sufficient evidence that more than 1.5% of the users experience flulike symptoms
b. since p - value > α, reject the null hypothesis and conclude that there is not sufficient evidence that more than 1.5% of the users experience flulike symptoms
c. since p - value < α, do not reject the null hypothesis and conclude that there is sufficient evidence that more than 1.5% of the users experience flulike symptoms
d. since p - value > α, do not reject the null hypothesis and conclude that there is not sufficient evidence that more than 1.5% of the users experience flulike symptoms

Explanation:

Step1: Recall the decision rule for hypothesis testing

In hypothesis testing, if the P - value is less than the significance level \(\alpha\) (\(P-\text{value}<\alpha\)), we reject the null hypothesis. If \(P - \text{value}>\alpha\), we do not reject the null hypothesis.

Step2: Compare the P - value and \(\alpha\)

We are given that \(\alpha = 0.1\) and \(P-\text{value}=0.209\). Since \(0.209>0.1\) (i.e., \(P - \text{value}>\alpha\)), we do not reject the null hypothesis.

Answer:

D. Since \(P-\text{value}>\alpha\), do not reject the null hypothesis and conclude that there is not sufficient evidence that more than \(1.5\%\) of the users experience flulike symptoms.