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 (alpha = 0.1) level of significance?
because (np_0(1 - p_0)=12.9>10), the sample size is less than 5% of the population size, and the sample can be reasonably assumed to be random, the requirements for testing the hypothesis are satisfied (round to one decimal place as needed.)
what are the null and alternative hypotheses?
(h_0:p = 0.015) versus (h_1:p>0.015) (type integers or decimals. do not round)
find the test statistic, (z_0).
(z_0=square) (round to two decimal places as needed)

Explanation:

Step1: Calculate the sample proportion $\hat{p}$

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 16$ (number of patients with flu - like symptoms) and $n=872$ (sample size). So, $\hat{p}=\frac{16}{872}\approx0.01835$.

Step2: Calculate the test statistic $z_0$

The formula for the test statistic in a one - sample proportion test is $z_0=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$. Here, $p_0 = 0.015$ (the proportion from competing drugs), $n = 872$, and $\hat{p}\approx0.01835$.
First, calculate the denominator: $\sqrt{\frac{0.015\times(1 - 0.015)}{872}}=\sqrt{\frac{0.015\times0.985}{872}}=\sqrt{\frac{0.014775}{872}}\approx\sqrt{0.00001694}\approx0.00412$.
Then, calculate the numerator: $\hat{p}-p_0=0.01835 - 0.015=0.00335$.
Now, $z_0=\frac{0.00335}{0.00412}\approx0.81$.

Answer:

$z_0\approx0.81$