Answer:
Yes, the confidence interval contradict this statement.
Step-by-step explanation:
The complete question is attached below.
The data provided is:
[tex]n_{1}=318\\n_{2}=297\\\bar x_{1}=25\\\bar x_{2}=24.1\\s_{1}=3.6\\s_{2}=3.8[/tex]
Since the population standard deviations are not provided, we will use the t-confidence interval,
[tex]CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\cdot s_{p}\cdot\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}[/tex]
Compute the pooled standard deviation as follows:
[tex]s_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}}=\sqrt{\frac{(318-1)(3.6)^{2}+(297-1)(3.8)^{2}}{318+297-2}}=2.9723[/tex]
The critical value is:
[tex]t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.05/2, (318+297-2)}=t_{0.025, 613}=1.962[/tex]
*Use a t-table.
The 95% confidence interval is:
[tex]CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\cdot s_{p}\cdot\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}[/tex]
[tex]=(25-24.1)\pm 1.962\times 2.9723\times \sqrt{\frac{1}{318}+\frac{1}{297}}\\\\=0.90\pm 0.471\\\\=(0.429, 1.371)\\\\\approx (0.43, 1.37)[/tex]
The 95% confidence interval for the difference between the mean weights is (0.43, 1.37).
To test the magazine's claim the hypothesis can be defined as follows:
H₀: There is no difference between the mean weight of 1-year old boys and girls, i.e. [tex]\mu_{1}-\mu_{2}=0[/tex].
Hₐ: There is a significant difference between the mean weight of 1-year old boys and girls, i.e. [tex]\mu_{1}-\mu_{2}\neq 0[/tex].
Decision rule:
If the confidence interval does not consists of the null value, i.e. 0, the null hypothesis will be rejected.
The 95% confidence interval for the difference between the mean weights does not consists the value 0.
Thus, the null hypothesis will be rejected.
Conclusion:
There is a significant difference between the mean weight of 1-year old boys and 1-year old girls.
