Suppose a marketing company randomly surveyed 259 households and found that in 150 of them, the woman made the majority of the purchasing decisions. Construct a 92% confidence interval for the population proportion of households where the women make the majority of the purchasing decisions. p'= 0.5791506 24/ 0.04 1.750686 Margin of Error: E = 28.17463 We are 92% confident that the proportion of households in the population where women make the majority of purchasing decisions is between x and X

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