Question

Question 5 of 10 (1 point) | Attempt 1 of Unlimited | View question in a popup 11.2 Section Exercise 14 (p-value, table) Hurricane damage: In August and September 2005, Hurricanes Katrina and Rita caused extraordinary flooding in New Orleans, Louisiana. Many homes were severely damaged or destroyed, and of those that survived, many required extensive cleaning. It was thought that cleaning of flood-damaged homes might present a health hazard due to the large amounts of mold present in many of the homes. In a sample of 360 residents Orleans Parish who had participated in cleaning of one or more homes, 76 had experienced symptoms of wheezing, and in a sample of 180 residents who had not participated in cleaning, 24 reported wheezing symptoms (numbers read from a graph). Can you conclude that the proportion of residents with wheezing symptoms is greater among those who participated in the cleaning of flood-damaged homes? Let P1 denote the proportion of residents with wheezing symptoms who had cleaned flood-damaged homes, and let p, be the population proportion with wheezing symptoms who did not participate in the cleaning of flood-damaged homes. Use the a=0.10 level of significance and the P-value method. O illa Part 1 of 5 State the null and alternate hypotheses. H: P1 = P2 U

Answer 1

The pooled proportion here is computed as:

P = (76 + 24) / (360 + 180) = 100/540 = 0.1852

The standard error here is computed as

$SE = \sqrt{P(1-P)(\frac{1}{n_1} + \frac{1}{n_2})} = \sqrt{0.1852(1 - 0.1852)(\frac{1}{360} + \frac{1}{180})}$

SE = 0.0355

The sample proportions now are computed here as:
p1 = 76/360 = 0.2111
p2 = 24/180 = 0.1333

Therefore the test statistic here is computed as:

$z^* = \frac{p_1 - p_2}{SE} = \frac{0.2111 - 0.1333}{0.0355} = 2.19$

Therefore 2.19 is the test statistic value here.

As this is a one tailed test, the p-value here is obtained from the standard normal tables as:
p = P(Z > 2.19) = 0.0143

As the p-value here is 0.0143 < 0.1 which is the level of significance here, therefore the test is significant here and we reject the null hypothesis here. Therefore we have sufficient evidence here that the first population proportion is more than the second.