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In: Statistics and Probability

Insurance companies are interested in knowing the population percent of drivers who always buckle up before...

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 415 drivers and find that 286 claim to always buckle up. Construct a 95% confidence interval for the population proportion that claim to always buckle up. Use interval notation, for example, [1,5]

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Expert Solution

We have given,          
          
x=286      
n=415      
Estimate for sample proportion
Level of significance is =1-0.95=0.05  
Z critical value(using Z table)=1.96

Confidence interval formula is          


=(0.6446,0.7337)         

Therefore, a 95% confidence interval for the population proportion that claim to always buckle up is   (0.6446,0.7337)  

orchestra answered 2 years ago
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