A large-sample 98 percent confidence interval for the proportion of hotel reservations that are canceled on theintended arrival day is (0.048, 0.112). What is the point estimate for the proportion of hotel reservations that are canceled on the intended arrival day from which this interval was constructed?(A) 0.032 (B) 0.064 (C) 0.080 (D) 0.160 (E) It cannot be determined from the information given.

Option C is correct.

The point estimate for the proportion of hotel reservations that are canceled on the intended arrival day from which this interval was constructed = 0.08

Explanation:

Confidence Interval = (Sample Mean) ± (Margin of error)

Upper limit of the confidence interval = (Sample Mean) + (Margin of error)

Lower limit of the confidence interval = (Sample Mean) - (Margin of error)

Let Sample mean = x

Margin of error = y

(0.048, 0.112) = x ± y

x + y = 0.112

x - y = 0.048

Solving this simultaneous equation,

2x = 0.112 + 0.048 = 0.16

x = (0.16/2) = 0.08

y = 0.112 - 0.08 = 0.032

Sample mean = 0.08

Margin of Error = 0.032

So, the point estimate for the proportion of hotel reservations that are canceled on the intended arrival day from which this interval was constructed = 0.08

Hope this Helps!!!

the answer is false?

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