Suppose That A Restaurant Launched A New Advertising Campaig
Suppose That A Restaurant Launched A New Advertising Campaign In Conju
Suppose that a restaurant launched a new advertising campaign in conjunction with a new menu. After one month, a sample of 250 bills was collected, with an average bill of 73.4 and a standard deviation of 38.6. The objective is to determine whether the new campaign and menu have increased the average bill above the previous average of 71.6 through a hypothesis test. The null hypothesis is that the population mean (μ) is less than or equal to 71.6, and since the population standard deviation is unknown, the sample standard deviation is used to perform a t-test.
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In the context of evaluating the effectiveness of a new advertising campaign combined with a new menu, it is critical to statistically analyze whether there has been a significant increase in the average restaurant bill. The hypothesis testing process provides a rigorous framework for making such an inference based on sample data. This analysis involves formulating the null and alternative hypotheses, calculating the appropriate test statistic, and interpreting the results to determine if there is enough evidence to support the claim that the campaign has increased spending.
Formulating the hypotheses
The null hypothesis (H₀) posits that the true average bill has not increased as a result of the campaign, expressed mathematically as H₀: μ ≤ 71.6. Conversely, the alternative hypothesis (H₁) suggests that the campaign has indeed increased the average bill, stated as H₁: μ > 71.6. Since the focus is on whether the average bill has risen above the previous average, this is a one-tailed test.
Sample data and assumptions
The sample size is n = 250, which is sufficiently large for the Central Limit Theorem to justify the approximation of the sampling distribution of the mean as normal, even with an unknown population standard deviation. The sample mean is 73.4, and the sample standard deviation (s) is 38.6. The previous average bill (μ₀) is 71.6. Our goal is to determine whether the observed sample mean provides enough evidence to reject the null hypothesis in favor of the alternative hypothesis.
Calculating the test statistic
The appropriate test statistic in this scenario is the t-statistic, given by:
t = (x̄ - μ₀) / (s / √n)
where x̄ is the sample mean, μ₀ is the hypothesized population mean, s is the sample standard deviation, and n is the sample size.
Applying the values
Substituting the values:
- x̄ = 73.4
- μ₀ = 71.6
- s = 38.6
- n = 250
The standard error (SE) is:
SE = s / √n = 38.6 / √250 ≈ 38.6 / 15.811 ≈ 2.441
The test statistic is then:
t = (73.4 - 71.6) / 2.441 ≈ 1.8 / 2.441 ≈ 0.737
Rounding to three decimal places, the test statistic is:
t ≈ 0.737
Conclusion
This test statistic can now be compared to critical t-values from the t-distribution with n - 1 = 249 degrees of freedom, or used to compute a p-value. Given the large sample size, the critical value for a one-tailed test at conventional significance levels (e.g., α=0.05) will be close to 1.645. Since 0.737
Implications
The statistical analysis suggests that although the average bill appears higher (73.4 vs. 71.6), this difference is not statistically significant at the conventional levels. As such, the restaurant management should consider additional data or other factors before concluding the campaign’s effectiveness in increasing customer spending.
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