Chapter 18 CW #4 (Paired t-test and t-interval)

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After seeing countless commercials claiming one can get cheaper car insurance from an online company, a local insurance agent was concerned that he might lose some customers. To investigate, he randomly selected profiles (car type, coverage, driving record, etc.) for 10 of his clients and checked online price quotes for their policies. The comparisons are shown in the table.

a) Is there evidence that drivers might save money by switching to the online company? (This is d = local – online; if we want to save with online, we would want our difference to be greater than for Ha)

b) How much of a difference would this be?

We're testing local price - online price, so our null hypothesis is that the mean of the differences is 0. Our alternate hypothesis is that the online company might be better, so the alternate hypothesis is that mean of the differences is positive, or greater than 0.

When we do STAT -> TEST -> t-test, the result is a p-value of .215.

If you want to do this by hand, use the formula t = xbar - μd / (s / sqrtn).

.215 is much larger than any reasonable alpha level, so we fail to reject the null hypothesis. There is insufficient evidence to believe that drivers might save money by switching to the online company.

b) We can now do a t-interval to determine our range of prices.

Do STAT -> TEST -> t-interval to yield an interval of (-79.79, 171.59). (This corroborates our failure to reject the null hypothesis, because 0 is included in the set.)

So the difference could be anywhere from a loss of $79.79 to a gain of $171.59.