Problem 7

Average (population) = 900

SD (population) = 1000

Average (sample) = 850

N (sample) = 2500

(a) Testing for statistical significance, we get:

SE = 1000/ sqrt (2500) = 20

Z = (900-850)/20 = 2.5

From the normal table, statistical power = 99.5% for Z (2.5), therefore it is statistically significant and we can conclude that implementing the hotline actually reduces health-care costs per family.

(b) P-value is 0.5% (i.e. 100 – 99.5) and can be interpreted as assuming that hotlines don’t help to reduce healthcare costs (null hypothesis is true), there would be a 0.5% chance of occurrence of getting average cost per family down to $850 or lower.

(c) No. Given that there is only one-tail and the histogram is negatively skewed, the p-value for the one-tail remains 0.5%. Additionally, we should be considering the one-tailed p-value since we are only interested in determining whether the helpline helped to decrease (and not increase) the average healthcare cost of $900.

Problem 12

Average (population) = 800

SD (population) = 1000

N (sample) = 100

Average (sample) = 1100

SE = 1000/sqrt (100) = 100

Z = (1100-800)/100 = 3

From the normal table, statistical power = 99.87% for Z (3), therefore it is statistically significant and we can conclude that the high per-patient average expense cannot be explained by chance and is a cause for concern due to some other problem.

Problem 20

No, it does not mean that the tripling the risk of cancer is significant. The analyst means that the increase in expected risk in the sample of 892 women doesn’t exceed the limit due to chance variation. The increase in expected risk by 3 times is consistent with chance variation and is something we would have expected to see in a sample of 892 women.

Problem 22

Treatment group: 1287 patients, 3.5% affected

Control group: 1299 patients, 1.92% affected

So, difference we saw between groups is 1.58%

SE for % treatment group = sqrt (3.5 x 96.5) / 1287) = 0.51%

SE for % control group = sqrt (1.92 x 98.08) / 1299) = 0.38%

So SE for difference = sqrt (0.51^2 + 0.38^2) = 0.64%

Variation of 1.58% represents 1.58/0.64 = 2.47 standard errors of variation. This is a statistically significant amount of variation beyond what is usually expected from chance variation, and therefore does go beyond the threshold of “statistical significance”.

Problem 34

Average (population) = 50

SD (population) = 10

N = 75 accounts

Average (sample) =