Question

a company wants to estimate the proportion of households in which its product would be purchased. A random sample of 600 households was selected. The results indicate that, 155 of the households would buy the product. The confidence interval of 90% of the population proportion of households that would buy the product is:   (0.67329; 0.88670)

WHY IS IT (0.67329; 0.88670)??

what table, graph or method did you use?

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explanation please

Answer 1

[Note: For the data you have given, your confidence interval answer is not correct. Please see below for all the steps and the correct answer.]

n = 600    

p = 155/600 = 0.258333333    

% = 90    

Standard Error, SE = √{p(1 - p)/n} =    √(0.258333333333333(1 - 0.258333333333333))/600 = 0.017869771

z- score = 1.644853627    

Width of the confidence interval = z * SE =     1.64485362695147 * 0.0178697706673506 = 0.02939316

Lower Limit of the confidence interval = P - width =     0.258333333333333 - 0.0293931570949827 = 0.22894018

Upper Limit of the confidence interval = P + width =     0.258333333333333 + 0.0293931570949827 = 0.28772649

The 90% confidence interval is [0.22894, 0.28773]

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