In: Statistics and Probability
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.
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 * ((( * (1 - )) / n)
= 1.96 (((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)