# Here are summary statistic for randomly selected weights of

Here are summary statistic for randomly selected weights of newborn girls: $$\displaystyle{n}={211},\ \overline{{{x}}}={33.7}{h}{g},\ {s}={6.8}{h}{g}.$$ Construct a confidence intercal estimate of the mean. Use a 90% confidence level. Are these results very different from the confidence interval $$\displaystyle{32.1}{h}{g}{ < }\mu{ < }{34.3}{h}{g}$$ with only 16 sample values, $$\displaystyle\overline{{{x}}}={33.2}{h}{g}$$, and $$\displaystyle{s}={2.6}{h}{g}$$?
What is the confidence interval for the population mean $$\displaystyle\mu$$?
Are the results between the two confidence intervals very different?
a) Yes, because the confidence interval limits are not similar.
b) No, because each confidence interval contains the mean of the other confidence interval.
c) No, because the confidence interval limits are similar
d) Yes, because one confidence interval does not contain the mean of the other confidence interval.

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habbocowji
Step 1
Solution:
From the given information, $$\displaystyle{n}={211},\ \overline{{{x}}}={33.7}{h}{g}$$ and $$\displaystyle{s}={6.8}{h}{g}.$$
Step 2
The confidence level is 0.90.
$$\displaystyle{1}-\alpha={0.90}$$
$$\displaystyle\alpha={1}-{0.90}$$
$$\displaystyle\alpha={0.10}$$
Then, a 90% confidence interval for the population mean is
$$\displaystyle\overline{{{x}}}\pm{t}_{{{\frac{{\alpha}}{{{2}}}},\ {n}-{1}}}{\left({\frac{{{s}}}{{\sqrt{{{n}}}}}}\right)}={33.7}\pm{t}_{{{\frac{{{0.10}}}{{{2}}}},\ {211}-{1}}}{\left({\frac{{{6.8}}}{{\sqrt{{{211}}}}}}\right)}$$
$$\displaystyle={33.7}\pm{t}_{{{0.05},\ {210}}}{\left({\frac{{{6.8}}}{{\sqrt{{{211}}}}}}\right)}$$
$$\displaystyle={33.7}\pm{\left({1.625}\right)}{\left({\frac{{{6.8}}}{{\sqrt{{{211}}}}}}{\left[\text{Using the excel function}\ ={T}.{I}{N}{V}{\left({0.05},\ {210}\right)}\right]}\right.}$$
$$\displaystyle={33.7}\pm{0.7734}$$
$$\displaystyle={\left({33.7}-{0.7734},\ {33.7}+{0.07734}\right)}$$
$$\displaystyle={\left({32.9},\ {34.5}\right)}$$
Thus, a 90% confidence interval for the population mean is $$\displaystyle{32.9}{h}{g}{ < }\mu{ < }{34.5}{h}{g}$$
Step 3
From the given information, the confidence interval for the population mean is $$\displaystyle{32.1}{h}{g}{ < }\mu{ < }{34.3}{h}{g}$$
From the above two confidence intervals it can be observed that both lower and upper limits of the two confidence intervals are different.
Step 4
Reason for the incorrect options:
From the two confidence intervals it can be observed that limits are not same. To determine the similarity of the two confidence intervals it has to the limits of the intervals.
Thus, the options B, C and D are incorrect.
###### Not exactly what you’re looking for?
Barbara Meeker

Step 1
Sample size $$\displaystyle{n}={211}$$
Sample mean $$\displaystyle\overline{{{x}}}={33.7}$$
Standard deviation $$\displaystyle{s}={6.8}$$
we use t distibution as population standard deviation is unknown.
degree of freedom $$\displaystyle={n}-{1}$$
$$\displaystyle={211}-{1}$$
$$\displaystyle={210}$$
Significance level, $$\displaystyle\alpha={0.10}$$
cratical value $$\displaystyle{t}_{{{c}}}={1.652}$$ (Use Excel Formula :T.INV(probability, $$deg_{freedom}$$)
90% Confidence interval $$\displaystyle=\overline{{{x}}}\pm{\left({t}_{{{c}}}\times{\frac{{{s}}}{{\sqrt{{{n}}}}}}\right)}$$
$$\displaystyle={33.7}\pm{\left({1.652}\times{\frac{{{6.8}}}{{\sqrt{{{211}}}}}}\right)}$$
$$\displaystyle={\left({32.927},\ {34.473}\right)}$$
The Two confidence intervals are (32.9, 34.5) and (32.1, 34.3)
The mean 33.2 is included in (32.9, 34.5)
The mean 33.7 is included in (32.1, 34.3)
So each confidence interval contains the mean of the other confidence interval.

karton

$$\begin{array}{} \text{Step 1} \\\text{Given:} \\\text{Sample size} (n)=211 \\\text{Sample mean} (\bar{x})=28.2 \\\text{Standard deviation} (s)=6.4 \\\alpha=0.05 \\(\alpha=0.05,\ dF=210) \\t_{2.10,\ 0.05}=\pm 1.97 \\\text{Confidence interval is} \\=\bar{x}\pm t\frac{s}{\sqrt{n}} \\=28.2\pm\frac{1.97(6.4)}{\sqrt{211}} \\=28.2\pm0.867 \\=(27.3,\ 29.067) \\\text{Therefore, Confidence interval} \end{array}$$
$$\begin{array}{} \\27.3<\mu<29.067 \\t=\frac{\bar{x}_{1}-\bar{x}_{2}}{\sqrt{\frac{s_{1}}{n_{1}}^{2}+\frac{s_{2}}{n_{2}}^{2}}} \\\bar{x}_{2}=27.6 \\s_{2}^{2}=2.1 \\n_{2}=19 \\t=\frac{28.2-27.6}{\sqrt{\frac{(6.4)^{2}}{211}+\frac{(2.1)^{2}}{19}}} \\=0.920 \\dF=211+19-2 \\=229 \\t_{229,\ 0.09}=1.97 \end{array}$$
Our t-statistic is less than 1.97. So,
We have to accept Noll Hypothesy
There is no difference brom the confidence interval brom the two Sample
$$27.3<\mu<29.067$$