Final answer:
To determine the probability that the result of the pharmacologic study could be due to chance, a hypothesis test needs to be performed. The observed proportion is compared to the expected proportion under the null hypothesis to calculate a z-score, which can then be used to find the corresponding p-value. If the p-value is less than the significance level, the null hypothesis can be rejected.
Step-by-step explanation:
To determine the probability that the result of the pharmacologic study could be due to chance, we need to perform a hypothesis test. The null hypothesis would state that the proportion of patients with blood-glucose levels at least 1.5 times the upper limit of normal is equal to the proportion expected by chance.
We can then calculate the z-score using the given information and find the corresponding p-value from the standard normal distribution. If the p-value is less than a predetermined significance level (usually 0.05), we can reject the null hypothesis and conclude that the result is unlikely due to chance.
Let's assume we set the significance level at 0.05 for this example. We can calculate the z-score using the formula:
z = (p - P) / sqrt(P(1-P) / n) where p is the observed proportion, P is the expected proportion under the null hypothesis, and n is the sample size. In this case, p = 76 / 5000 = 0.0152 (approximately), P = 0.05, and n = 5000.
Plugging in these values, we get: z = (0.0152 - 0.05) / sqrt(0.05(1-0.05) / 5000)
= -30.056
We can then find the p-value corresponding to this z-score using a standard normal distribution table or a statistical calculator. The p-value in this case is extremely close to 0, indicating that the result is highly unlikely to be due to chance.
Therefore, we can reject the null hypothesis and conclude that the result is statistically significant.