Can you answer these statistics problems?

Question

I need to know how I go about solving this problem and what the answer is. I don't even understand what is being asked!

You want to compare the yearly costs of auto insurance companies. You select a sample of 15 families (some with one and some with multiple drivers). To make the data comparable, certain features, such as the amount deductible and limits of liability, are standardized. The sample information is reported below. At the .10 significance level, can we conclude that there is a difference in the amounts quoted?

FamilyAmerican Insurance /St. Paul Insurance

Becker$2090-$1610
Berry1683-1247
Cobb1402-2327
Debuck1830-1367
DuBrul930-1461
Eckroate697-1789
German1741-1621
Glasson1129-1914
King1018-1956
Kucic1881-1772
Meredith1571-1375
Obeid874-1527
Price1579-1767
Phillips1577-1636
Tresize860-1188

Answer 1

The sample size is small (n=15), so for precision you need to use a t-statistic and, also, an unbiased estimator of variance. Yes, you assume that the differences come from a normal distribution with mean zero (that's your null hypothesis). You compute the sample mean of the difference x_i = quote1-quote2, for i = 1,2...15. This is

xbar = (1/15)[(2090-1610) + (1683-1247) + (1402-2327) +...].

You compute the sample variance s^2 of x. This is the average square deviation from the sample mean:

s^2 = (1/14)Sum(x_i - xbar)^2 (sum is over i = 1,2...15)

Note the division is by n-1 = 14 (not 15); with a small n, this can matter. Now you compute the t-statistic (where s is of course the root of s^2)

t = (xbar - 0) / (s/sqrt(n)) = xbar sqrt(15) / s.

If x_i are normal and independent, this t has Student's distribution (which becomes normal as n increases, but n=15 is too small to claim normality of t). Your test is two-tailed (the alternative hypothesis is that the quotes are different w/o prejudging which one is higher). So you compare the t-value with the 95-th percentile and the 5-th percentile of Student's t-distribution with n-1 = 14 degrees of freedom. The 95-th percentile is 1.761 (symmetrically, the 5-th percentile is -1.761). This means that under the null hypothesis (that the two quotes are the same on average), the probability of t being more than 1.761 in absolute value is exactly 10% (split half and half between the low and the high values). So if your computed value of t is either below -1.761 or above 1.761, you reject the null hypothesis and conclude that the two companies give different quotes on average.
Now just compute:

xbar = -158.53
s = 535.57
t = -158.53*3.873/535.57 = -1.1462

so, even at the not very demanding 10% level, there is not enough evidence to reject the hypothesis that the quotes are on average the same.

Answer 2

The question is simpler than it looks I think.

Basically, each family is getting insurance quotes from two insurance companies. On average, you want to know if one of those companies is cheaper. That means, you look at the difference of the two quotes.

You hypothesis is that the two companies are the same; i.e., that the average is 0!!!

What you do is calculate the average and standard deviation of these difference of the two quotes. Then, you assume normal distribution and use your tables to find out whether you average is close enough to zero or not. If its not, you reject the hypothesis and conclude one is cheaper than the other.

Good luck.

Now,

Answer 3

the first learn team that took a million (325mg) buffered aspirin pill the different day, below a million% or .ninety 4% suffered heart attacks. the 2d learn team that took a million placebo the different day, a million.seventy one% suffered heart attacks. This finally ends up in the decision that the first learn team suffered 40 5% a lot less heart attacks than the 2d learn team that took the placebo. This also finally ends up in the very undeniable truth that taking a million (325mg) buffered aspirin the different day very much reduces the prospect for a heart attack.