A local donut shop wishes to test out its donuts against those of the competition which is planning to set up in the community. Each of 50 subjects tastes 2 unidentified donuts and says which they perfer. 31 or 62% of the subjects perfer the local company's donuts to the competition's donuts. Test the claim that a majority(over 50%) of people prefer the local company donuts at the significance level of 0.01. What is your practical conclusion? Do a full hypotheses test.
Suggest how you would design and conduct this experiment.
Using a formula for a binomial proportion one-sample z-test with your data included, we have:
z = .62 - .50 / √[(.50)(.50)/50]
Finish the calculation.
Use a z-table to find the critical or cutoff value at 0.01 for a one-tailed test.
If the z-test statistic calculated above exceeds the critical value from the z-table, reject the null. If the z-test statistic does not exceed the critical value from the z-table, do not reject the null.
I hope this will help get you started.
To test the claim that a majority of people prefer the local company's donuts, we can conduct a hypothesis test using the given data. Here is a step-by-step explanation of how to perform the test.
1. Formulate the hypotheses:
- Null Hypothesis (H₀): p ≤ 0.5 (Less than or equal to 50% of people prefer the local donuts)
- Alternative Hypothesis (H₁): p > 0.5 (More than 50% of people prefer the local donuts)
2. Choose the significance level (α) at 0.01. This is the maximum probability of rejecting the null hypothesis when it is true.
3. Count the number of subjects who prefer the local donuts. In this case, 31 subjects preferred the local donuts.
4. Calculate the test statistic. Since the sample size is large (50) and we are dealing with proportions, we can use the normal distribution approximation. The test statistic (Z) can be calculated using the formula:
Z = (p̂ - p₀) / √[(p₀ * (1 - p₀)) / n]
where p̂ is the sample proportion, p₀ is the hypothesized proportion in the null hypothesis, and n is the sample size.
p̂ = 31/50 = 0.62 (proportion of subjects who prefer the local donuts)
p₀ = 0.5 (proportion in the null hypothesis)
n = 50 (sample size)
Plugging in these values, we get:
Z = (0.62 - 0.5) / √[(0.5 * (1 - 0.5)) / 50] ≈ 1.39
5. Determine the critical region. Since the alternative hypothesis is one-tailed (p > 0.5), we need to find the critical Z-value for a significance level of 0.01. Looking up the value in a Z-table or using statistical software, we find the critical Z-value to be approximately 2.33 (for a one-tailed test).
6. Compare the test statistic to the critical value. Since 1.39 < 2.33, it falls within the non-rejection region.
7. Make a decision. Since the test statistic falls within the non-rejection region, we fail to reject the null hypothesis. There is not enough evidence to conclude that a majority of people prefer the local donuts.
Based on this hypothesis test, we cannot conclude that a majority of people prefer the local donuts. The practical conclusion suggests that the local donut shop's claim of having a majority preference is not supported by the data. To conduct this experiment, you would need to randomly select the 50 subjects and have them taste two unidentified donuts (one from the local shop and one from the competition) and record their preferences.