Question

In: Statistics and Probability

A restaurant wants to estimate the average bill paid by a party of two for dinner...

A restaurant wants to estimate the average bill paid by a party of two for dinner (service hours after 5 p.m.) They take a random sample of 100 bills and find that they average $80 with a standard deviation of $40. In making an 80% confidence interval for the parameter of interest, they should get a margin of error of about

± $80

± $50

± $10

± $5

A random sample of 900 homeowners in Centre County is taken in order to estimate the average amount of money spent by homeowners on landscaping this spring. It turns out that this group spent an average of $200 with a standard deviation of $150. A 95% confidence interval for the average amount spent on landscaping by all Centre County homeowners is calculated to be $200 ± $10. Which is true about this situation?

The true average amount spent must be in this interval.

The procedure used here produces intervals that cover the right answer for 95% of all samples.

This confidence interval won’t cover the right answer since a histogram of the amount spent is skewed.

The interval would be larger if we used an 80% confidence level instead of 95%.

Two polls about the President’s policy for the War in Syria used the same methodology and had the same sample size but had different question wording. In the first poll 45% of those surveyed favored the policy, while in the second poll 55% favored the policy. True or False: even though the question wording might have changed the results, it would not have any affect on the margin-of-error.

True

False

Expert Solution

Q 1) The Margin of error is

Answer: ± $5

Q 2) Answer: The true average amount spent must be in this interval.

Q 3) Answer: True

orchestra answered 2 years ago

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