Question

in a clinical trial, 28 out of 888 patients taking a prescription drug daily complained of flulike symptoms. suppose that it is known that 2.8% of the patients taking competing drugs complain of flulike symptoms. is there sufficient evidence to conclude that more than 2.8% of this drug's user experience flulike symptoms as a side effect at the aplha 0.05 level of significance

Answer 1

The p-value (0.0001) is less than α = 0.05. Based on this, we should reject the null hypothesis.

What is a null hypothesis?

A null hypothesis (H₀) can be defined the opposite of an alternate hypothesis (H₁) and it asserts that two (2) possibilities are the same.

How to calculate value of the test statistic?

The test statistic can be calculated by using this formula:

$t=\frac{x\;-\;u}{\frac{\delta}{\sqrt{n} } }$

Where:

• x is the sample mean.

• u is the mean.

• is the standard deviation.

• n is the number of hours.

For this clinical trial (study), we should use a t-test and the null and alternative hypotheses would be given by:

H₀: p = 0.028

H₁: p > 0.028

For the sample proportion, we have:

Sample proportion, P = 28/888

Sample proportion, P = 0.032.

Next, we would calculate the t-test as follows:

$t=\frac{0.032\;-\;0.028}{\frac{0.02 \times 0.028}{888 } }\\\\t=\frac{0.004}{\sqrt{\frac{0.00056}{888 } }}$

t = 0.004/0.00079

t = 5.06.

For the p-value, we have:

P-value = P(t > 5.06)

P-value = 1 - P(t < 5.06)

P-value = 1 - 0.9999

P-value = 0.0001.

Therefore, the p-value (0.0001) is less than α = 0.05. Based on this, we should reject the null hypothesis.

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