Math Problem Statement

You are the store manager of Brand Factory in Mumbai. In the past, you observed that customer arrivals and purchases go up significantly during the sales season compared to regular days in your store. At the same time, it is observed that the customers get irritated while standing at the billing counter. As a store manager, you have counters and necessary actions like involving more staff, opening additional billing counters, and managing the crowd during the sales season. Irrespective of all the arrangements, the customers remain dissatisfied. You wanted to know if the customer's irritation is due to the duration of waiting time. You want to check if the waiting time at the billing counter has changed during the sales season to that of the previous month's average waiting time of 4.5 minutes and a standard deviation of 1.2 minutes. To resolve this business problem, you selected a sample of 35 case points at the billing counter for 1 hour (The data is given in the annexure below). Use this sample information to determine whether there is evidence at the 0.05 significance level that the population mean waiting time at the billing counter has changed in the sales season from the previous month's average waiting time of 4.5 minutes.

Solution

To determine whether the population mean waiting time at the billing counter has changed during the sales season compared to the previous month's average of 4.5 minutes, we can perform a hypothesis test. Specifically, we'll use a one-sample tt-test, given that we have a sample mean and standard deviation but do not know the population standard deviation exactly.

Hypothesis Setup

  • Null Hypothesis (H0H_0): The mean waiting time during the sales season is equal to the previous month's average waiting time. H0:μ=4.5 minutesH_0: \mu = 4.5 \text{ minutes}

  • Alternative Hypothesis (H1H_1): The mean waiting time during the sales season is different from the previous month's average waiting time. H1:μ4.5 minutesH_1: \mu \neq 4.5 \text{ minutes}

Significance Level

The significance level (α\alpha) is 0.05.

Test Statistic

The test statistic for a one-sample tt-test is calculated as follows: t=xˉμ0snt = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} Where:

  • xˉ\bar{x} is the sample mean
  • μ0\mu_0 is the hypothesized population mean (4.5 minutes)
  • ss is the sample standard deviation
  • nn is the sample size

Degrees of Freedom

The degrees of freedom (dfdf) for the tt-test is n1n - 1.

Decision Rule

Compare the calculated tt-statistic to the critical tt-value from the tt-distribution table at df=34df = 34 (since n=35n = 35) for a two-tailed test at α=0.05\alpha = 0.05. If t|t| is greater than the critical value, we reject the null hypothesis.

Steps to Solve

  1. Calculate the sample mean (xˉ\bar{x}) and sample standard deviation (ss) from the data provided.
  2. Compute the tt-statistic using the formula above.
  3. Determine the critical tt-value from the tt-distribution table for df=34df = 34 at α=0.05\alpha = 0.05.
  4. Compare the tt-statistic to the critical value and make a decision.

You can provide the sample data for the next steps, and I'll help you calculate the necessary values and conclude the hypothesis test.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
One-sample t-test
Sampling

Formulas

t-test formula: t = (x̄ - μ₀) / (s / √n)

Theorems

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Suitable Grade Level

College Level