The mean +1 sd of In [calcium intake (mg)] among 25 females, 12 to 14 years of age, below the poverty level is 6.56 + 0.64. Similarly, the mean + 1 sd

Emeli Hagan 2021-02-09 Answered
The mean +1 sd of In [calcium intake (mg)] among 25 females, 12 to 14 years of age, below the poverty level is 6.56+0.64.
Similarly, the mean + 1 sd of In [calcium intake (mg)] among 40 females, 12 to 14 years of age, above the poverty level is 6.80+0.76.
1) Test for a significant difference between the variances of the two groups.
2) What is the appropriate procedure to test for a signifi- cant difference in means between the two groups?
3) Implement the procedure in Problem 8.3 using the critical-value method.
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Expert Answer

Benedict
Answered 2021-02-10 Author has 108 answers
Given:
n1=40,s1=0.76
n2=25,s2=0.64
1) Hypothesis
H0:σ1=σ2
H1:σ1σ2
Test statistics would be
F=s12s22
=0.7620.642
=1.41
Degree of freedom of numerator 401=39
Degree of freedom of denominator 251=24
P-value of the test =0.3759
Therefore P-value is greater than 0.05, therefore it is fail to reject the null hypothesis.
Conclusion is that there is no significant difference between the populations.
2) For this two sample t-test is used the reason is that population variances are equal.
3)
H0:μ1=μ2
H1:μ1μ2
Degree of freedom
=n1+n22
=40+252
=63
For two tailed test critical values ±1.998
If t is less than -1.998 or greater than 1.998, fall to reject the null hypothesis.
Pooled standard deviation
Sp=(n11)s12+(n21)s22n1+n22
=(401)0.762+(251)0.646240+252
=0.71666
And the t-statistics would be
t=x2x2Sp1n1+1n2
=6.86.560.71666×140+125
=1.3136
By seeing the t value, it is not lying in the rejection region, therefore it is fail to reject the null hypothesis.
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