Suppose You Have A Product You Sell In Two Markets

Question

Suppose You Have A Product Which You Sell In Two Markets Of Comparable Suppose you have a product which you sell in two markets of comparable size and you want to test the hypothesis that the demand for the product is the same in the two markets. In practice, you test that the quantity demanded is the same and find that for a sample of 17 different time periods with different prices, the average quantity demanded in the first market is 1765 and in the second market it is 1688. The standard deviation of the differences is 60.5. What is the test statistic? Answer to two decimal places.

Dr. Jack HW Helper · Accepted Answer

The problem presented involves testing whether the demand for a product is statistically similar across two comparable markets using sample data. This is a classic hypothesis testing scenario involving the comparison of two population means. The specific test to be used here is the two-sample t-test for the difference between means, which is appropriate when the standard deviations are estimated from sample data and the population variances are unknown. Given the data, the key parameters are as follows: - Sample size (n): 17 time periods for each market. - Sample mean demand in market 1 ($\bar{x}_1$): 1765 - Sample mean demand in market 2 ($\bar{x}_2$): 1688 - Difference between sample means ($\bar{x}_1 - \bar{x}_2$): 77 - Standard deviation of the differences ($s_d$): 60.5 The hypothesis test aims to evaluate: - Null hypothesis ($H_0$): There is no difference in mean demand between the two markets ($\mu_1 = \mu_2$), or equivalently, $\mu_d = 0$ - Alternative hypothesis ($H_A$): There is a difference ($\mu_1 \neq \mu_2$) The test statistic for the paired differences follows a t-distribution, given by: $$ t = \frac{\bar{d} - \mu_{d0}}{s_d / \sqrt{n}} $$ where: - $\bar{d}$: the mean difference ($\bar{x}_1 - \bar{x}_2 = 77$) - $\mu_{d0}$: the hypothesized mean difference under the null (0) - $s_d$: standard deviation of differences (60.5) - $n$: number of pairs (17) Substituting the values: $$ t = \frac{77 - 0}{60.5 / \sqrt{17}} $$ Calculating the denominator: $$ 60.5 / \sqrt{17} = 60.5 / 4.1231 \approx 14.676 $$ Calculating the t-statistic: $$ t = 77 / 14.676 \approx 5.25 $$ Therefore, the test statistic is approximately 5.25, to two decimal places. References Field, A. (2013). Discovering Statistics Using IBM SPSS Statistics. Sage Publications. Newbold, P., Carlson, W. L., & Thorne, B. (2013). Statistics for Business and Economics. Pearson. Zimmerman, J. (2012). Business Statistics: A First Course. McGraw-Hill Education. Agresti, A., &

Suppose You Have A Product Which You Sell In Two Markets Of Comparable

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