A personal computer company launched an advertising campaign in the hopes of boosting sales. A random sample of 35 days before the advertising blitz showed mean sales of 120 computers per day with a standard deviation of 29. A random sample of 30 days after the advertisements showed mean sales of 130 computers per day with a standard deviation of 35. Test whether the advertising campaign is statistically effective at the .05 level of significance. (FYI tdata= -1.24) A. Null: $\mu_1 \ge \mu_2$ Alt: $\mu_1 < \mu_2$ B. $t_{data} = \frac{\bar{x_1} - \bar{x_2}}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} = \frac{(120 - 130)}{\sqrt{\frac{29^2}{35} + \frac{35^2}{30}}} = -1.24$ C. $dF =$
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05 The null hypothesis is that the mean sales before the advertising campaign is greater than or equal to the mean sales after the campaign ($\mu_1 \ge \mu_2$). The alternative hypothesis is that the mean sales before the campaign is less than the mean sales after Show more…
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A personal computer company wishes to increase sales with an advertising campaign. The first random sample of 15 days was taken before the campaign and had a mean sales of 120 computers per day with a standard deviation of 30. The second random sample of 15 days was taken after the campaign and had a mean sales of 125 computers per day with a standard deviation of 35. Write the hypotheses to be tested. A) H0: μ1 = μ2, H1: μ1 < μ2 B) H0: μ1 = μ2, H1: μ1 < μ2 C) H0: μ1 ≥ μ2, H1: μ1 < μ2 D) H0: μ1 < μ2 E) H0: σ1 = σ2 F) Ha: σ1 ≠ σ2
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Sales at a fast-food restaurant average $6,000 per day. The restaurant decided to introduce an advertising campaign to increase daily sales. In order to determine the effectiveness of the advertising campaign, a sample of 49 day's sales were taken. The sample showed average daily sales of $6,300. From past history, the restaurant knew that its population standard deviation is about $1,000. If the level of significance is 0.01, have sales increased as a result of the advertising campaign? a) Fail to reject the null hypothesis. b) Reject the null hypothesis and conclude the mean is higher than $6,000 per day. c) Reject the null hypothesis and conclude the mean is lower than $6,000 per day. d) Reject the null hypothesis and conclude that the mean is equal to $6,000 per day.
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