The Coffee Institute has claimed that more than 40 percent of American adults regularly have a cup
of coffee with breakfast. A random sample of 450 individuals revealed that 200 of them were regular
coffee drinkers at breakfast. What is the prob value for a test of hypotheses seeking to show that the
Coffee Institute’s claim was correct? (Hint: Test H0: p = 0.4, versus H1: p > 0.4)
To determine the probability value for a test of hypotheses seeking to show that the Coffee Institute's claim was correct, we can use statistical hypothesis testing. In this case, we want to compare the proportion of American adults who regularly have a cup of coffee with breakfast to the claimed proportion of more than 40 percent.
Let's define the null hypothesis (H0) and the alternative hypothesis (H1):
H0: p = 0.4 (The proportion of American adults regularly having coffee with breakfast is 40%)
H1: p > 0.4 (The proportion of American adults regularly having coffee with breakfast is greater than 40%)
We have a random sample of 450 individuals, of which 200 are regular coffee drinkers at breakfast. To calculate the probability value (also known as p-value), we can use a one-sample proportion test using the normal approximation to the binomial distribution.
We need to calculate the test statistic, which is the standard normal test statistic (z-score) calculated as:
z = (p̂ - p0) / sqrt(p0 * (1 - p0) / n)
where p̂ is the sample proportion (200/450), p0 is the hypothesized proportion under the null hypothesis (0.4), and n is the sample size (450).
Substituting the values:
z = ((200/450) - 0.4) / sqrt(0.4 * (1 - 0.4) / 450)
z = (0.4444 - 0.4) / sqrt(0.4 * 0.6 / 450)
z = 0.0444 / sqrt(0.24 / 450)
z = 0.0444 / sqrt(0.000533333)
z ≈ 0.0444 / 0.0231
z ≈ 1.9236
Next, we need to calculate the probability of obtaining a z-score as extreme as 1.9236 or greater. This probability is typically obtained from a standard normal distribution table or by using statistical software.
Using a standard normal distribution table, the probability value associated with a z-score of 1.9236 is approximately 0.0276 (or 2.76%).
Therefore, the probability value for this test of hypothesis is approximately 0.0276, suggesting that there is evidence to reject the null hypothesis.