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4. A genetics researcher performed a study to determine whether a relationship e

ID: 3158058 • Letter: 4

Question

4. A genetics researcher performed a study to determine whether a relationship exists between gender and blood type. He obtained the following sample data presented in the two-way table. Blood Type A O B AB

a. [5] Conduct the appropriate hypothesis test indicated by this scenario to determine whether the geneticist can conclude a relationship exists between gender and blood type. You may use technology, but make sure you include all important information (hypotheses, test statistic with degrees of freedom if appropriate, p-value, statistical conclusion using = 0.05, and practical conclusion.)

b. [2] Use the results from your calculator to give the expected counts for this scenario if the null hypothesis is true (i.e. there is no relationship.) You may round the values to the nearest whole number on this document.

Type A O B AB FEMALE 115 141 27 13 MALE 86 91 26 9

Explanation / Answer

Question 4

Part a

Solution:

Here, we have to check the claim or hypothesis whether there is any relationship exists between the gender and blood type. The null and alternative hypothesis is given as below:

Null hypothesis: H0: The variables gender and blood type are independent.

Alternative hypothesis: Ha: The average gender and blood type are not independent.

The level of significance or alpha value is given as 0.05.

The test statistic formula is given as below:

Chi square = [(O – E)^2 / E]

Degrees of freedom = (r – 1)*(c – 1)

Where r is the number of rows and c is the number of columns.

Here, r = 2 and c = 4

d.f. = (2 – 1)(4 – 1) = 1*3 = 3

The test is given as below:

Chi-Square Test

Observed Frequencies

Blood Group

Gender

A

O

B

AB

Total

Female

115

141

27

13

296

Male

86

91

26

9

212

Total

201

232

53

22

508

Expected Frequencies

Blood Group

Gender

A

O

B

AB

Total

Female

117.1181

135.1811

30.88189

12.8189

296

Male

83.88189

96.8189

22.11811

9.181102

212

Total

201

232

53

22

508

Data

Level of Significance

0.05

Number of Rows

2

Number of Columns

4

Degrees of Freedom

3

Results

Critical Value

7.814728

Chi-Square Test Statistic

1.867376

p-Value

0.600384

Do not reject the null hypothesis

Here, we get the p-value as 0.6 which is greater than the given level of significance or alpha value 0.05, so we do not reject the null hypothesis that the variables gender and blood type are independent.

Part b

Solution:

The expected counts rounded to nearest whole number is given as below:

Expected Frequencies

Blood Group

Gender

A

O

B

AB

Total

Female

117

135

31

13

296

Male

84

97

22

9

212

Total

201

232

53

22

508

Chi-Square Test

Observed Frequencies

Blood Group

Gender

A

O

B

AB

Total

Female

115

141

27

13

296

Male

86

91

26

9

212

Total

201

232

53

22

508

Expected Frequencies

Blood Group

Gender

A

O

B

AB

Total

Female

117.1181

135.1811

30.88189

12.8189

296

Male

83.88189

96.8189

22.11811

9.181102

212

Total

201

232

53

22

508

Data

Level of Significance

0.05

Number of Rows

2

Number of Columns

4

Degrees of Freedom

3

Results

Critical Value

7.814728

Chi-Square Test Statistic

1.867376

p-Value

0.600384

Do not reject the null hypothesis

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