Respuesta :
Answer:
Step-by-step explanation:
1.
From the given information;
sample size n = 10
[tex]\bar x = \dfrac{\sum x}{n}[/tex]
we can have the weight(x) and the additional column [tex](x - \bar x)^2[/tex] to be as follow;
weight (x) [tex](x - \bar x)^2[/tex]
12.32 0.805
11.87 0.200
12.84 0.841
11.48 0.003
12.66 1.530
8.51 8.486
14.13 9.328
12.95 2.332
9.34 4.339
8.63 7.801
[tex]\sum x = 114.23[/tex] [tex]\sum(x- \bar x)^2 = 33.664[/tex]
[tex]\bar x = \dfrac{\sum x}{n}[/tex]
[tex]\bar x = \dfrac{114.23}{10} \\ \\ \bar x= 11.423[/tex]
Sample standard deviation
[tex]s = \sqrt {\dfrac{\sum(x-\bar x)^2}{n-1}} \\ \\s = \sqrt {\dfrac{33.664}{10-1}} \\ \\ s= \sqrt{\dfrac{33.664}{9}}\\ \\ s= 1.934[/tex]
degree of freedom df = n-1
= 10 -1
= 9
For 90% confidence interval C.I ∝ = 1 - 0.90 = 0.10
∝/2 = 0.10/2 = 0.05
From t-distribution table;
the value of t having an area of ( ∝/2 = 0.05) in its upper tail & for (df = 9) is given by:
[tex]t_{0.05} = 1.833[/tex]
The Margin of error ;
M.O.E = [tex]t_{0.05} * \dfrac{s}{\sqrt{n}}[/tex]
M.O.E = [tex]1.833 * \dfrac{1.934}{\sqrt{10}}[/tex]
M.O.E = 1.121
90% C.I for mean weight μ
The lower limit of 90% C.I = 11.423 - 1.121 = 10.302
The upper limit of 90% C.I = 11.423 + 1.121 = 12.544
Therefore, 90% C.I for the mean weight of the baby girls is; 10.302 < μ < 12.544.
B.
The mean weight of 13.9 pounds of two month old baby is outside the limits of 90% Confidence Interval C.I as derived above ;
Therefore; we conclude that it is not reasonable to believe that the mean weight of two-month-old baby girls may be the same as that of two-month-old baby boys.
2.
weight (x) [tex](x- \bar x) ^2[/tex]
4.753 0.193
4.374 0.004
4.175 0.019
4.676 0.131
4.424 0.0121
4.228 0.007
4.124 0.0361
4.254 0.004
3.955 0.129
4.196 0.014
4.291 0.000529
[tex]\sum x = 47.45[/tex] [tex]\sum (x- \bar x)^2 = 0.549729[/tex]
where sample size n = 11
[tex]\bar x = \dfrac{\sum x}{n}[/tex]
[tex]\bar x = \dfrac{47.45}{11}[/tex]
[tex]\bar x = 4.314[/tex]
sample standard deviation
[tex]s = \sqrt {\dfrac{\sum(x-\bar x)^2}{n-1}} \\ \\s = \sqrt {\dfrac{0.549729}{11-1}} \\ \\ s= \sqrt{\dfrac{0.549729}{10}}\\ \\ s= 0.235[/tex]
degree of freedom = n - 1 = 11 - 1 = 10
For 80% confidence interval C.I ∝ = 1 - 0.80 = 0.20
∝/2 = 0.20/2
∝/2 = 0.1
From t-distribution table;
the value of t having an area of ( ∝/2 = 0.1) in its upper tail & for (df = 10) is given by:
[tex]t_{0.1} = 1.37[/tex]
The Margin of error ;
M.O.E = [tex]t_{0.05} * \dfrac{s}{\sqrt{n}}[/tex]
M.O.E = [tex]1.37 * \dfrac{0.235}{\sqrt{11}}[/tex]
M.O.E = 0.0971
80% C.I for mean weight μ
The lower limit of 80% C.I = 4.314 - 0.0971 = 4.2169
The upper limit of 80% C.I = 4.314 + 0.0971 = 4.4111
Therefore, 80% C.I for the mean rate is; 4.2169 < μ < 4.4111.
