A research manager at Coca-Cola claims that the true proportion, p, of cola drinkers that prefer
Coca-Cola over Pepsi is greater than 0.50. In a consumer taste test, 100 randomly selected people
were given blind samples of Coca-Cola and Pepsi. 58 of these subjects preferred Coca-Cola. Is there
sufficient evidence at the 5% level of significance to validate Coca-Cola’s claim? Conduct an
appropriate hypothesis test using (i) the p-value method and (ii) the critical value method.
Null hypothesis:
Ho: p = .50 -->meaning: population proportion is equal to .50
Alternative hypothesis:
Ha: p > .50 -->meaning: population proportion is greater than .50
Using a formula for a binomial proportion one-sample z-test with your data included, we have:
z = .58 - .50 -->test value (58/100 = .58) minus population value (.50)
divided by
√[(.50)(.50)/100]
Using a z-table, find the critical or cutoff value for a one-tailed test (upper tail) at .05 level of significance. The test is one-tailed because the alternative hypothesis is showing a specific direction (greater than).
The p-value will be the actual level of the test statistic. You can use a z-table to determine that value.
I hope this will help get you started.
To determine whether there is sufficient evidence to validate Coca-Cola's claim, we need to conduct a hypothesis test. Let's perform this test using both the p-value method and the critical value method.
(i) The p-value method:
1. State the null hypothesis (H₀) and the alternative hypothesis (H₁):
H₀: p ≤ 0.50 (Coca-Cola preference is not greater than 0.50)
H₁: p > 0.50 (Coca-Cola preference is greater than 0.50)
2. Calculate the test statistic (z-score):
The test statistic for a proportion can be calculated using the formula:
z = (x - np₀) / √(np₀(1 - p₀))
where:
x = number of successes (58)
n = sample size (100)
p₀ = hypothesized proportion under the null hypothesis (0.50)
Plugging in the values, we have:
z = (58 - 100*0.50) / √(100*0.50*(1-0.50))
= 8 / √25
= 8 / 5
= 1.6
3. Determine the p-value:
The p-value is the probability of obtaining a test statistic as extreme as the one calculated (z = 1.6 or higher) under the null hypothesis. P(Z ≥ 1.6) can be found using a standard normal distribution table or a statistical calculator.
The p-value for z = 1.6 is approximately 0.0548.
4. Compare the p-value to the significance level (α):
We are given a significance level of 5% or 0.05. Since the p-value (0.0548) is greater than the significance level, we fail to reject the null hypothesis.
Conclusion: There is not sufficient evidence at the 5% level of significance to validate Coca-Cola's claim that the true proportion of cola drinkers who prefer Coca-Cola is greater than 0.50.
(ii) The critical value method:
1. State the null hypothesis (H₀) and the alternative hypothesis (H₁):
Same as in the p-value method.
2. Set the significance level (α):
We are given a significance level of 5% or 0.05.
3. Calculate the critical value:
The critical value (zc) is obtained from the standard normal distribution table or a statistical calculator for the given significance level.
For a one-tailed test with a significance level of 0.05, the critical value for rejection is approximately 1.645.
4. Calculate the test statistic (z-score):
Same as in the p-value method.
5. Compare the test statistic to the critical value:
If the test statistic (z = 1.6) is greater than the critical value (zc = 1.645), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Since 1.6 < 1.645, we fail to reject the null hypothesis.
Conclusion: There is not sufficient evidence at the 5% level of significance to validate Coca-Cola's claim that the true proportion of cola drinkers who prefer Coca-Cola is greater than 0.50.
Note: In both methods, if the p-value or the test statistic is less than the significance level (α), we would reject the null hypothesis and conclude that there is sufficient evidence to support Coca-Cola's claim.