Since the level of significance is not provided assuming it to
be 5%.
(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be
tested:
H0: The two variables are independent i.e. having a cell phone is
independent of one’s grade level
Ha: The two variables are dependent i.e. having a cell phone is
not independent of one’s grade level
This corresponds to a Chi-Square test of independence.
Degrees of Freedom
The number of degrees of freedom is df = (2 - 1) * (2 - 1) =
1
Critical value and Rejection Region
Based on the information provided, the significance level is
α=0.05, the number of degrees of freedom is df = (2 - 1) * (2 - 1)
= 1, so the critical value is 3.8415.
Then the rejection region for this test becomes
R={χ2:χ2>3.8415}.
(2)Test Statistics
The Chi-Squared statistic is computed as follows:
(3)P-value
The corresponding p-value for the test is
p=Pr(χ2>0.5282)=0.4673
(4)Decision about the null hypothesis
Since it is observed that χ2=0.5282<χ2_crit=3.8415, it is then
concluded that the null hypothesis is NOT rejected.
(5)Conclusion
It is concluded that the null hypothesis Ho is NOT rejected.
Therefore, there is NOT enough evidence to claim that the two
variables having a cell phone and grade level are dependent, at the
0.05 significance level.
Conditions:
a. The sampling method is simple random sampling.
b. The data in the cells should be counts/frequencies
c. The levels (or categories) of the variables are mutually
exclusive.
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