Question

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 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 340 residents of Orleans Parish who had participated in the cleaning of one or more homes, 69 had experienced symptoms of wheezing, and in a sample of 172 residents who had not participated in the cleaning, 26 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 P, denote the proportion of residents with wheezing symptoms who had cleaned flood-damaged homes and p2 denote the proportion of residents with wheezing symptoms who had not cleaned flood-damanged homes. Use the a= 0.05 level of significance and the P-value method with the TI-84 Plus calculator Part: 0 / 4 Part 1 of 4 State the null and alternate hypotheses. Ho: H: This hypothesis test is a (Choose one) 7 test.

a. State the null and alternate hypotheses.
b. Compute the P-value.
C. Determine whether to reject Ho.
D. State a conclusion.

Answer 1

Solution:

Given:

Sample 1:

n₁ = Number of residents of Orleans Parish who had participated in the cleaning of one or more homes = 340

x₁ = Numbers residents experienced symptoms of wheezing = 69

Sample 2:

n₂ = Number of residents of Orleans Parish who had not participated in the cleaning homes = 172

x₂ = Numbers residents experienced symptoms of wheezing = 26

We have to test the proportion of residents with wheezing symptoms is greater among those who participated in the cleaning of flood-damaged homes.

Level of significance = 0.05

Part 1) State the null and alternative hypotheses.

H₀: p₁ = p₂

H₁: p₁ > p₂

Part 2) Compute the P-value.

Use following steps in TI 84 plus calculator:

Press STAT and select TESTS

Under TESTS , select 2-PropZTest

TI-84 Plus Silver Edition TEXAS INSTRUMENTS EDIT CALC TESTS 1: Z-Test... 2:T-Test... 3:2-SampZTest.... 4:2-SampTTest.... 5:1-PropZTest... 382-PropZTest.... 7+Interval... STAT PLOTF1 TBLSETF2 FORMAT F3 CALC F4 TABLE F5 WINDOW ZOOM TRACE GRAPH 2ND CU MODE DEL ALPHA X.T.. STAT MATH APPS PRAWC PRGM VARS CLEAR

Enter Numbers:

TI-84 Plus Silver Edition TEXAS INSTRUMENTS 2-PropZTest X1:69 ni: 340 X2:26 n2: 172 P1:#P2 P2 PZ Calculate Draw STAT PLOTF1 TBLSETF2 FORMAT F3 CALC F4 TABLE F5 Y WINDOW ZOOM TRACE GRAPH 2ND QUIT MODE INS DEL ALDO ALPHA TESTA MATH LINK X.T.@.n ANGLE APPS LIST STAT DRAWC PRGM DISTR VARS CLEAR

Click on Calculate and press Enter

TI-84 Plus Silver Edition TEXAS INSTRUMENTS 2-PropTest P1P2 Z=1.423499566 P=.0772957409 1=.2029411765 A2=.1511627907 =. 185546875 STAT PLOTF1 TBLSETF2 FORMAT F3 CALC F4 TABLE F5 Y WINDOW ZOOM TRACE GRAPH 2ND QUE MODE INS DEL ADGE ALPHA TESTA MATH MA LIST X.T..n STAT ANGLEB DRAWC APPS PRGM DISTR VARS CLEAR

P-value = 0.0772957

P-value = 0.0773

(Round final answer to specified number of decimal places)

Part 3) Determine whether to reject H0.

Decision Rule:
Reject null hypothesis H0, if P-value < 0.05 level of significance, otherwise we fail to reject H0.

Since P-value = 0.0773 > 0.05 level of significance, we fail to reject H0.

Part 4) State a conclusion.

At 0.05 level of significance, we do not have sufficient evidence to conclude that: the proportion of residents with wheezing symptoms is greater among those who participated in the cleaning of flood-damaged homes.