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 9Explanation / 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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