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 s

Kyran Hudson

Kyran Hudson

Answered question

2021-03-12

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.
8.2 Test for a significant difference between the variances of the two groups.
8.3 What is the appropriate procedure to test for a signifi- cant difference in means between the two groups?
8.4 Implement the procedure in Problem 8.3 using the critical-value method.
8.5 What is the p-value corresponding to your answer to Problem 8.4?
8.6 Compute a 95% Cl for the difference in means between the two groups.

Answer & Explanation

krolaniaN

krolaniaN

Skilled2021-03-13Added 86 answers

Step a
Since you haven't specified which section you're seeking for, we're only allowed to respond to the first three subparts of your question at once. If you have questions about the remaining subparts, re-post them separately.
Given: 
n1=40,s1=0.76 
n2=25,s2=0.64 
Step b
8.2 
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 
Because the P-value is higher than 0.05, the null hypothesis cannot be rejected.
The conclusion is that the populations do not significantly differ from one another.
Step c 
8.3 
The reason the t-test with two samples is used is that the population variances are equal.
Step d 
8.4 
H0:μ1=μ2 
H1:μ1μ2 
Degree of freedom 
=n1+n22 
=40+252 
=63 
For two tailed test critical values ±1.998 
Fall to reject the null hypothesis if t is not more than 1.998 or less than -1.998.
Pooled standard deviation 
Sp=(n11)S12+(n21)S22n1+n22 
=(401)0.762+(251)0.64240+252 
=0.71666 
And the t-statistics would be 
t=x1x2Sp1n1+1n2 
=6.86.560.71666×140+125 
=1.3136 
The null hypothesis cannot be rejected since the t value does not fall within the rejection zone.

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