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 95% confidence interval to estimate the population proportion is .

The 95% confidence interval to estimate the population proportion is (0.2365, 0.3635).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

For this problem, we have that:

95% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval to estimate the population proportion is (0.2365, 0.3635).

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step-by-step explanation: ; )

its 4 because 2 times 2 is 4, if you double the number two you get four