Question 8 (1 point) A restaurant wants to test a new in-store marketing scheme

Question

Question 8 (1 point) A restaurant wants to test a new in-store marketing scheme in a small number of stores before rolling it out nationwide. The new ad promotes a premium drink that they want to increase the sales of. 6 locations are chosen at random and the number of drinks sold are recorded for 2 months before the new ad campaign and 2 months after. The average difference in the sales quantity (after- before) is -118.963 with a standard deviation of 30.679. When calculating a 90% confidence interval to estimate the true difference in nationwide sales quantity before the ad campaign and after, what is the margin of error? 1) 12.5246 2) 18.0324 3) 2.7132 4) 25.2378 5) 24.3377

Explanation / Answer

DF = 6 - 1 = 5

With 5 degrees of freedom and 90% confidence interval the critical value is t0.05, 5 = 2.015

Margin of error = t0.05, 5 * S/sqrt (n)

= 2.015 * 30.679/sqrt(6)

= 25.2372

Option-4 is the correct answer

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