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A large trucking company wants to estimate the proportion of its tracker truck population with refrigerated carrier capacity. A random sample of 200 tracker trucks is taken and 30% of the sample have refrigerated carrier capacity. The 90% confidence interval to estimate the population proportion is _______.

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Answer:

Step-by-step explanation:

Given that a large trucking company wants to estimate the proportion of its tracker truck population with refrigerated carrier capacity.

Sample size n =200

Sample proportion p = 0.30

q = 1-p =0.70

Std error = [tex]\sqrt{\frac{pq}{n} } \\=0.0324[/tex]

For 90% confidence interval, Z critical = ±1.645

Hence margin of error = ±1.645(0.0324) =±0.0533

Confidence interval = 0.30±0.0533

= (0.2467, 0.3533)

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