Last Month The Waiting Time At The Drive-Through Window

Last Month The Waiting Time At The Drive Through Window Of A Fast F

Last month the waiting time at the drive through window of a fast-food restaurant was 3.7 minutes. The franchise has installed a new process intended to reduce waiting time. After the installation, a random sample of 64 orders is selected. The sample mean waiting time is 3.57 minutes with a sample standard deviation of 0.8 minutes. At a 5% level of significance, is there sufficient evidence that the population mean waiting time is now less than 3.7 minutes? What is your conclusion?

In a survey of 1,040 adults conducted by American Express Incentive Services, 563 responded that they would prefer being given $100 rather than a day off from work. At a 5% level of significance, is there enough evidence from this survey to conclude that more than half of all adults would rather have $100 than a day off? What is your conclusion? What is the p-value for this test?

A tire manufacturer produces tires that are believed to have a mean life of at least 25,000 miles when the production process is working correctly. Based on past experience, the population standard deviation of the lifetime of the tires is 3,500 miles. Assume a level of significance for testing of 5%, and a random sample of 100 tires:

A) What would be the consequences of making a Type II error in this problem?

B) Compute the probability of making a Type II error if the true population mean is 24,000 miles. Please Note: A hypothesis test answer must contain: a Null and an Alternate Hypothesis, a computed value of the test statistic, a critical value of the test statistic, a Decision, and a Conclusion.

Paper For Above instruction

Analysis of Hypothesis Testing Scenarios in Business and Manufacturing

Hypothesis testing is a fundamental aspect of statistical analysis used to make inferences about population parameters based on sample data. It provides a structured framework to test assumptions, assess evidence, and guide decision-making in various contexts such as business operations, consumer preferences, and manufacturing quality control. This paper discusses three distinct scenarios: a change in service efficiency at a fast-food drive-through, consumer preferences from survey data, and tire durability in manufacturing, each illustrating different applications of hypothesis testing principles.

Scenario 1: Evaluating Change in Drive-Through Wait Times

In the first scenario, a fast-food franchise records an average wait time of 3.7 minutes before implementing a new process aimed at reducing waiting duration. Post-implementation, a random sample of 64 orders yields a mean of 3.57 minutes with a standard deviation of 0.8 minutes. The question at hand is whether there is statistical evidence to support that the mean waiting time has decreased significantly from 3.7 minutes at a 5% significance level.

Null hypothesis (H₀): μ = 3.7 minutes (the mean wait time is unchanged)

Alternative hypothesis (H_a): μ

Given the large sample size, the z-test for the population mean is appropriate, especially since the standard deviation from the sample is used as an estimate. The test statistic is calculated as:

z = (x̄ - μ₀) / (σ / √n) = (3.57 - 3.7) / (0.8 / √64) = -0.13 / (0.8 / 8) = -0.13 / 0.1 = -1.3

The critical value for a one-tailed test at α = 0.05 is approximately -1.645. Since the computed z-value (-1.3) is greater than -1.645, we fail to reject the null hypothesis.

Therefore, at the 5% significance level, there is not enough evidence to conclude that the mean waiting time has decreased significantly. The data suggests that the reduction might be due to random variation rather than a true effect.

Scenario 2: Analyzing Consumer Preference Data

The second scenario involves survey data where 563 out of 1,040 adults prefer receiving $100 over a day off, indicating a proportion p̂ = 563/1040 ≈ 0.5413. The hypothesis test assesses whether more than half of the adult population prefers the monetary incentive.

Null hypothesis (H₀): p = 0.5

Alternate hypothesis (H_a): p > 0.5

The test statistic for a proportion is:

z = (p̂ - p₀) / √(p₀(1 - p₀) / n) = (0.5413 - 0.5) / √(0.5 * 0.5 / 1040) ≈ 0.0413 / √(0.25 / 1040) ≈ 0.0413 / 0.0155 ≈ 2.66

The critical value for a one-tailed test at α = 0.05 is approximately 1.645. Since 2.66 > 1.645, we reject the null hypothesis, indicating there is statistically significant evidence that more than half of all adults prefer $100.

The p-value corresponding to z = 2.66 is approximately 0.0039, reinforcing the conclusion that the preference for $100 is significantly greater than 50%.

Scenario 3: Tire Durability and Type II Error Analysis

The third scenario assesses the manufacturing process for tires, where the specified mean lifespan is at least 25,000 miles with a population standard deviation of 3,500 miles. A sample of 100 tires is taken to test this claim.

• Part A: Consequences of a Type II Error
• A Type II error occurs when the null hypothesis (H₀: μ ≥ 25,000 miles) is falsely accepted, ignoring a true alternative. In this context, failing to detect that the mean tire life is less than 25,000 miles could lead to continued production of substandard tires, resulting in increased customer dissatisfaction, warranty costs, and potential damage to the manufacturer’s reputation. This can also cause long-term economic losses due to product recalls and reduced market share.
• Part B: Calculating the Probability of Type II Error (β)
• Assuming the true mean is 24,000 miles, the goal is to compute the probability that the test fails to reject H₀.
• Null hypothesis: H₀: μ = 25,000 miles
• Alternative hypothesis: H_a: μ
• The test statistic for the z-test is:
• Z = (x̄ - μ₀) / (σ / √n) = (x̄ - 25,000) / (3,500 / √100) = (x̄ - 25,000) / 350
• The critical z-value for a one-tailed test at α=0.05 is -1.645. The corresponding critical sample mean (x̄_c) is:
• x̄_c = μ₀ + z_critical (σ / √n) = 25,000 + (-1.645) 350 ≈ 25,000 - 575.75 ≈ 24,424.25 miles
• Now, calculating the probability that the sample mean is above this critical value if the true mean is 24,000 miles:
• Standardizing x̄ = 24,424.25 with μ = 24,000 miles:
• Z = (24,424.25 - 24,000) / 350 ≈ 424.25 / 350 ≈ 1.21
• Using standard normal tables, the probability of observing a z-value less than 1.21 (since we are concerned with the probability of failing to reject H₀) is approximately 0.8869. Therefore, the probability of a Type II error is about 88.69%, indicating a high chance of missing a truly defective batch if the mean drops to 24,000 miles.
• Conclusion
• These three scenarios demonstrate the importance of hypothesis testing in decision-making across industries. Whether evaluating process improvements, understanding consumer behavior, or ensuring product quality, statistical inference allows organizations to assess evidence systematically and make informed choices. Paying attention to types of errors and p-values helps mitigate risks associated with incorrect conclusions, ultimately supporting better operational and strategic outcomes.
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