Question

PLEASE EXPLAIN HOW TO SOLVE IT WITH EXCEL

1)

A researcher wanted to determine which method would help smokers quit smoking. He divided a group of 183 smokers into three groups: Group 1 is given a patch, Group 2 is given oral medication, and Group 3 has no intervention. The results of the study are in the below table.

	Group 1	Group 2	Group 3
Did Not Quit	52	54	51
Quit	13	9	4

*Source: Harris Poll

1. Determine the null and alternative hypotheses.
   • H0:H0: The methods for helping people quit smoking are the same.
     Ha:Ha: The methods for helping people quit smoking follow a different distribution.
   • H0:H0:The methods for helping people quit smoking are independent.
     Ha:Ha: The methods for helping people quit smoking are dependent.
   
   
2. Determine the test statistic. Round your answer to two decimal places.
   χ2 = Incorrect3.96
   
3. Determine the p-value. Round your answer to four decimal places.
   p-value =
   
4. Make a decision.
   • Fail to reject the null hypothesis.
   • Reject the null hypothesis.
   
   
5. Make a conclusion.
   • There is sufficient evidence to support the claim that the methods for helping people quit smoking are different.
   • There is not sufficient evidence to support the claim that the methods for helping people quit smoking are different.

2)

Pain medications have side effects where one common side effect is dizziness. A researcher wanted to determine whether some pain medications produce more or less dizziness than others. The results from his study are shown below in the table. Apply a test for homogeneity to determine whether the proportion of people taking a pain medication within each treatment group that experience dizziness are the same at the α=0.01α=0.01 level of significance.

	Tordol	Placebo	Naproxen	Tylenol	Advil
Dizziness	83	32	36	5	8
No dizziness	4,063	1,832	1,330	382	337

1. Determine the null and alternative hypotheses.
   • H0:H0:The proportion of people within each treatment group who experienced dizziness are independent.
     Ha:Ha: The proportion of people within each treatment group who experienced dizziness are dependent.
   • H0:H0: The proportion of people within each treatment group who experienced dizziness are the same.
     Ha:Ha: The proportion of people within each treatment group who experienced dizziness are different.
   
   
2. Determine the test statistic. Round your answer to two decimal places.
   χ2=χ2=
   
3. Determine the pp-value. Round your answer to four decimal places.
   pp-value =
   
4. Make a decision.
   • Fail to reject the null hypothesis.
   • Reject the null hypothesis.
   
   
5. Make a conclusion.
   • There is not sufficient evidence to support the claim that the proportion of people within each treatment group who experienced dizziness are different.
   • There is sufficient evidence to support the claim that proportion of people within each treatment group who experienced dizziness are different.

Answer 1

1.

H₀:The methods for helping people quit smoking are independent.
H_a: The methods for helping people quit smoking are dependent.

Observed frequencies:

	Group 1	Group 2	Group 3	Totals
Did not quit	52=O11	54=O12	51=O13	157=O10
Quit	13=O21	9=O22	4=O23	26=O20
Totals	65=O01	63=O02	55=O03	183=N

Expected frequencies:

Eij 0000; N

	Group 1	Group 2	Group 3
Did not quit	55.765=E11	54.0492=E12	47.1858=E13
Quit	9.235=E21	8.9508=E22	7.8142=E23

$Following~table~shows~\frac{(O_{ij}-E_{ij})^2}{E_{ij}}:$

0.2542	0	0.3083
1.5349	0.0003	1.8618

$Value~of~\chi^2=\sum_{i=1}^2\sum_{j=1}^3\frac{(O_{ij}-E_{ij})^2}{E_{ij}}=3.96\\\\ P-value=P(\chi^2>3.96|\chi^2\sim \chi^2_2)=0.1381\\\\ (R~code:~round(1-pchisq(3.96,2),4))$

Since P-value>0.05, Option: Fail to reject the null hypothesis.

Conclusion: There is not sufficient evidence to support the claim that the methods for helping people quit smoking are different.

2.

H₀: The proportion of people within each treatment group who experienced dizziness are the same.
H_a: The proportion of people within each treatment group who experienced dizziness are different.

Observed frequencies:

	Tordol	Placebo	Naproxen	Tylenol	Advil	Totals
Did not quit	83=O11	32=O12	36=O13	5=O14	8=O15	164=O10
Quit	4,063=O21	1,832=O22	1,330=O23	382=O24	337=O25	7,944=O20
Totals	4146=O01	1864=O02	1366=O03	387=O04	345=O05	8108=N

Expected frequencies:

Eij 0000; N

	Tordol	Placebo	Naproxen	Tylenol	Advil
Did not quit	83.8609=E11	37.703=E12	27.63=E13	7.8278=E14	6.9783=E15
Quit	4062.139=E21	1826.297=E22	1338.37=E23	379.1722=E24	338.0217=E25

$Following~table~shows~\frac{(O_{ij}-E_{ij})^2}{E_{ij}}:$

	0.0088	0.8626	2.5355	1.0215	0.1496
	0.0002	0.0178	0.0523	0.0211	0.0031

$Value~of~\chi^2=\sum_{i=1}^2\sum_{j=1}^3\frac{(O_{ij}-E_{ij})^2}{E_{ij}}=4.67 \\\\ P-value=P(\chi^2>4.67|\chi^3\sim \chi^2_3)=0.1976\\\\ (R~code:~round(1-pchisq(4.67,3),4))$

Since P-value>0.01, Option: Fail to reject the null hypothesis.

Conclusion: There is not sufficient evidence to support the claim that the proportion of people within each treatment group who experienced dizziness are different.