A marketing company survey found that, in the 100 households surveyed, in 65 households the women...

A marketing company survey found that, in the 100 households surveyed, in 65 households the women made the majority of the purchasing decisions. Find a 95% confidence interval for the proportion of households where women made the majority of purchasing decisions. Round your answer to the nearest hundredth.

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Expert Solution

Solution :

Given that,

Point estimate = sample proportion = = x / n = 65/100=0.65

1 - = 1- 0.65 =0.35

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z 0.025 = 1.96 ( Using z table )

Margin of error = E = Z/2 * $\sqrt$ ((( * (1 - )) / n)

= 1.96 ( $\sqrt$ ((0.65*0.35) /100 )

E = 0.093

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.65-0.093 < p < 0.65+0.093

(0.557 , 0.743)

orchestra answered 1 year ago

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