The installation of a radon abatement device is recommended in any home where the mean radon concentration is 4.0 picocuries per liter (pci/l) or more, because it is thought that long-term exposure to sufficiently high doses of radon can increase the risk of cancer. seventy-five measurements are made in a particular home. the mean concentrationwas 3.72 pci/l, and the standard deviation was 1.93 pci/l. a. the home inspector who performed the test says that since the mean measurement is less than 4.0, radon abatement is not necessary. explain why this reasoning is incorrect. b. because of health concerns, radon abatement is recommended whenever it is plausible that the mean radon concentration may be 4.0 pci/l or more. state the appropriate null and alternate hypotheses for determining whether radon abatement is appropriate. c. compute the p-value. would you recommend radon abatement? explain.

Step-by-step explanation:

a)  While the mean is below 4 the standard deviation tells us that there is a pretty high chance for the value to be above 4.  One standard deviation away is 1.79 - 5.65.  We are concerned with things over 4 so we'll look at the upper half, which is 3.72 - 5.65.  Being one standard deviation to the right means there is a 34.1% chance of the readings being in this range.  And there is even chances of it being slightly higher, though that is comparatively low.  But even as low as 10% is usually considered too high a chance to risk.  If you don't understand how the standard deviation got me those percents let me know.

b) alternative hypothesis is always the option where we want to prove it.  So we want to prove the concentration is 4 or above.  So the null is less than 4 and the alternative is greater than or equal to 4.  Do you know the correct symbols?  if not I can get those written out.  As for the p value we need the confidence level for the question, do you have that?

c-4 is the final answer! : )

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)