80% growth 20% growth
40% decay
40% growth
60% growth
20% decay
60% decay
80% decay
40(0.2)t = 1.6
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80(1.6)t = 20
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60(0.8)t = 1.4
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20(0.6)t = 1.2
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1.2(1.4)t = 80
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Arecent study focused on the number of times men and women who live alone buy take-out dinner in a month. assume that the distributions follow the normal probability distribution and the population standard deviations are equal. the information is summarized below. statistic men women sample mean 24.85 21.33 sample standard deviation 5.54 4.93 sample size 34 36 at the 0.01 significance level, is there a difference in the mean number of times men and women order take-out dinners in a month? state the decision rule for 0.01 significance level: h0: μmen= μwomen h1: μmen ≠ μwomen. (negative amounts should be indicated by a minus sign. round your answers to 3 decimal places.) compute the value of the test statistic. (round your answer to 3 decimal places.) what is your decision regarding the null hypothesis? what is the p-value? (round your answer to 3 decimal places.)