Question

researchers selected 872 patients at random among those who take a certain widely - used prescription drug daily in a clinical trial, 24 out of the 872 patients complained of flulike symptoms. suppose that it is known that 2.4% of patients taking competing drugs complain of flulike symptoms. is there sufficient evidence to conclude that more than 2.4% of this drugs users experience flulike symptoms as a side effect at the α = 0.1 level of significance?
because ( n p _ { 0 } ( 1 - p _ { 0 } ) = square square 10 ), the sample size is ( square 5 % ) of the population size, and the patients in the sample ( square ) selected at random, all of the requirements for testing the hypothesis ( square ) satisfied (round to one decimal place as needed)
what are the null and alternative hypotheses?
( h _ { 0 } square square square ) versus ( h _ { 1 } square square square )
(type integers or decimals. do not round.)
find the test statistic, ( z _ { 0 } )
( z _ { 0 } = square ) (round to two decimal places as needed.)
find the p - value.
p - value ( = square ) (round to three decimal places as needed.)
interpret the results
since the p - value is ( square ) than ( alpha ) ( square ) the null hypothesis. there ( square ) sufficient evidence at the ( alpha = square ) level of significance to conclude that ( square % ) of the users who take the prescription drug daily complained of flulike symptoms.

Explanation:

Step1: Check sample size conditions

Given \(n = 872\), \(p_0=0.024\)

$$ np_0(1 - p_0)=872\times0.024\times(1 - 0.024)=872\times0.024\times0.976 = 20.4 $$

Since \(np_0(1 - p_0)=20.4\gt10\), the sample size is likely less than \(5\%\) of the population size (as it's a clinical - trial sample, population of drug - users is large), and the patients are selected at random. So all requirements for testing the hypothesis are satisfied.

Step2: State null and alternative hypotheses

The null hypothesis \(H_0\) is a statement of no change or equality. The alternative hypothesis \(H_1\) is the claim we want to test.
Since we want to test if more than \(2.4\%\) (\(p_0 = 0.024\)) of patients experience flu - like symptoms, \(H_0:p = 0.024\) versus \(H_1:p>0.024\)

Step3: Calculate the sample proportion \(\hat{p}\)

\(\hat{p}=\frac{x}{n}\), where \(x = 24\) and \(n = 872\)

$$ \hat{p}=\frac{24}{872}\approx0.0275 $$

Step4: Calculate the test statistic \(z_0\)

The formula for the test statistic in a one - sample proportion test is \(z_0=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\)
Substitute \(\hat{p}=0.0275\), \(p_0 = 0.024\), and \(n = 872\)

$$ z_0=\frac{0.0275 - 0.024}{\sqrt{\frac{0.024\times(1 - 0.024)}{872}}}=\frac{0.0035}{\sqrt{\frac{0.024\times0.976}{872}}}=\frac{0.0035}{\sqrt{\frac{0.023424}{872}}}=\frac{0.0035}{\sqrt{0.00002686}}=\frac{0.0035}{0.00518}\approx0.68 $$

Step5: Calculate the P - value

Since \(H_1:p>p_0\) (right - tailed test), the P - value is \(P(Z>z_0)\)
Using the standard normal distribution table or a calculator, \(P(Z > 0.68)=1 - P(Z\leq0.68)\)
From the standard normal table, \(P(Z\leq0.68)=0.7517\)
So \(P - value=1 - 0.7517 = 0.248\)

Step6: Interpret the results

Since \(\alpha=0.1\) and \(P - value = 0.248>0.1\)
We do not reject the null hypothesis.

Answer:

• \(np_0(1 - p_0)=20.4\)
• \(H_0:p = 0.024\) versus \(H_1:p>0.024\)
• \(z_0\approx0.68\)
• \(P - value = 0.248\)
• Since the \(P\) - value (\(0.248\)) is greater than \(\alpha(0.1)\), we do not reject the null hypothesis. There is not sufficient evidence at the \(\alpha = 0.1\) level of significance to conclude that more than \(2.4\%\) of the users who take the prescription drug daily complained of flu - like symptoms.