In a survey conducted to determine, among other things, the cost of vacations, 64 individuals were...

In a survey conducted to determine, among other things, the cost of vacations, 64 individuals were randomly sampled. Each person was asked to compute the cost of her or his most recent vacation. The sample showed a sample mean of $1810. Assuming that the population standard deviation σ is $600, construct a 90% confidence interval for the average cost of all vacations. (Round your final answer to 3 decimal places).

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = $ 1810

Population standard deviation = = $ 600

Sample size = n = 64

At 90% confidence level

= 1 - 90%

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_0.05= 1.645

Margin of error = E = Z/2 * ( / $\sqrt$ n)

= 1.645 * ( 600 / $\sqrt$ 64 )

= 123.375

At 90% confidence interval estimate of the population mean is,

- E < < + E

1810 - 123.375 < < 1810 + 123.375

$ 1686.625 < < $ 1933.375

milcah answered 1 year ago

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