Please Explain. And write clearly, as this is also for me to learn and try under
ID: 3365298 • Letter: P
Question
Please Explain. And write clearly, as this is also for me to learn and try understand.
APPLICATIONS Auto Insurance The cost of Data set 10.40 EX1040 automobile insurance has become because the rates are dependent on so such as the province in which you live cars you insure, and the company with wb insured. The table is adapted from a Fraser Institute Digital Publication, Februa a sore subjet the number of which you are remiums in 10 provinces, 2004-2005.14 Province Avg. Earned PremiumAvg. Earned Premium in 205ed in 2004 BC ON SK MB NB AB 0C NL NS $1404 $1374 $1197 $1152 $1044 $1036 $988 $947 $871 $825 $1374 $1396 $1174 $1157 $1121 $1126 $983 $1014 $883 $847Explanation / Answer
a.
These pairs of observations are dependent because they are the avg earned premium from the same states for two different years. The current year avg earned premium may be dependent on the previous year avg earned premium.
b.
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: d = 0 (Mean difference in average premiums of 2004 and 2005 is 0)
Alternative hypothesis: d 0 (Mean difference in average premiums of 2004 and 2005 is not 0)
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).
The difference of the avg premium of 2004 and 2005 are
d = -30, 22, -23, 5, 77, 90, -5, 67, 12, 22
The mean difference is 23.7
The standard deviation of difference is, s = 41.49
SE = s / sqrt(n) = 41.49 / [ sqrt(10) ] = 13.12
DF = n - 1 = 10 -1 = 9
t = [ (x1 - x2) - D ] / SE = (d - D)/ SE = (23.7 - 0)/13.12 = 1.81
where d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 9 degrees of freedom is more extreme than 1.81; that is, less than -1.81 or greater than 1.81.
We use the t Distribution Calculator to find P(t < -1.81) = 0.0519, and P(t > 1.81) = 0.0519. Thus, the P-value = 0.0519 + 0.0519 = 0.1038.
Interpret results. Since the P-value (0.1038) is greater than the significance level (0.01), we cannot reject the null hypothesis and conclude that at significance level of 0.01, there is no significant evidence that the avg premium of year 2004 and 2005 are different.
c.
The P-value is 0.1038. This is the probability that a t statistic having 9 degrees of freedom is more extreme than 1.81; that is, less than -1.81 or greater than 1.81.
d.
t statistic for 99% confidence interval and df = 9 is 3.25
SE = 13.12
So, 99% confidence interval of difference in avg premiums is
(23.7 - 13.12 * 3.25, 23.7 + 13.12 * 3.25)
(-18.94, 66.34)