Q2-10BSC - Elementary Statistics

Question

Birth Weights of Boys Use these summary statistics for birth weights of 195 boys: \({\bf{\bar x}}\,{\bf{ = 32}}{\bf{.7}}\,{\bf{hg}}\),\({\bf{s}}\,{\bf{ = 6}}{\bf{.6}}\,{\bf{hg}}\). (based on Data Set 4 “Births” in Appendix B). Use a 95%  confidencelevel. Are the results very different from those found in Exercise 9? Does it appear that boys and girls have very different birth weights?

Step-by-Step Solution

Answer

The 95% confidence interval for birth weights of newborn boys is\(31.8\,{\rm{hg}} < \mu  < 33.6\,{\rm{hg}}\).                                                              

 

The results do not differ very much.

 

The birth weights of girls and boys do not differ very much.

Step 1: Given information

Based on Data set 4 “Birth” in Appendix B,the summary statistics for randomly selected weights of newborn girls as, \({\rm{n}}\,{\rm{ = }}\,195\), \(\bar x\, = 32.7\,{\rm{hg}}\), \({\rm{s}}\, = 6.6\,{\rm{hg}}\)

 

The confidence level is 95%.

Step 2: Describe confidence interval

A confidence interval is an estimate of interval that may contain true value of a population parameter. It is also known as interval estimate.

 

The general formula for confidence interval estimate of mean is,

\({\rm{Confidence}}\,\,{\rm{Interval}} = \left( {\bar x - {\rm{E}},\bar x + {\rm{E}}} \right)\,\,\,\,\,\,\,\,...\left( 1 \right)\)\(\)

Where, E is the margin of error, which is calculated as,

\({\rm{E}} = {t_{\frac{\alpha }{2}}} \times \frac{{\rm{s}}}{{\sqrt {\rm{n}} }}\)

Step 3: Find the appropriate distribution

If \(\sigma \) is not known and \({\rm{n}} > 30\) then t-distribution is suitable to find the confidence interval.

In this case,\(\sigma \)is unknown and \({\rm{n}} = 195\) which means n>30.So, thet-distribution applies here.

Step 4: Find critical value

To find the critical value \({t_{\frac{\alpha }{2}}}\), it requires a value for the degrees of freedom.

The degree of freedom is, 

\(\)

\(\begin{array}{c}{\rm{degree}}\,\,\,{\rm{of}}\,\,\,{\rm{freedom}}\,\,\, = n - 1\\ = 195 - 1\\ = 194\end{array}\)

 

The 95% confidence level corresponds to\(\alpha  = 0.05\), so there is an area of 0.025 in each of the two tails of the t-distribution.

 

Referring to Table A-3 critical value of t-distribution, the critical value of\({{\bf{t}}_{\frac{{\bf{\alpha }}}{{\bf{2}}}}}{\bf{ = }}{{\bf{t}}_{{\bf{0}}{\bf{.025}}}}\) is obtained from the intersection of  column with 0.05 for the “Area in Two Tails” (or use the same column with 0.025 for the “Area in One Tail”)and the row value number of degrees of freedom is 194 , which is 1.972.

Step 5: Find margin of error

The margin of error is calculated as,

\(\begin{array}{c}{\rm{E}} = {t_{\frac{\alpha }{2}}} \times \frac{{\rm{s}}}{{\sqrt {\rm{n}} }}\\ = 1.972 \times \frac{{6.6}}{{\sqrt {195} }}\\ = 0.9322\end{array}\)

Step 6: Find confidence interval

The confidence interval is obtained by substituting the value of margin of error in equation (1) as,

 

\(\begin{array}{c}{\rm{Confidence}}\,\,{\rm{Interval}} = \left( {\bar x - {\rm{E}},\bar x + {\rm{E}}} \right)\\ = \left( {32.7 - 0.9320\,\,,\,\,32.7 + 0.9320} \right)\\ = \left( {31.768\,\,,\,\,33.632} \right)\end{array}\)

 

Thus, the 95% confidence interval for estimate mean is \(31.8\,{\rm{hg}} < \mu  < 33.6\,{\rm{hg}}\).

Step 7: Compare the results

The confidence interval found in the Exercise-9 with 205 sample values for mean birth weights of girls is\(29.4\,{\rm{hg}} < \mu  < 31.4\,{\rm{hg}}\).

 

The confidence interval for estimate mean with 195 sample values for mean birth weights of boys is \(31.8\,{\rm{hg}} < \mu  < 33.6\,{\rm{hg}}\).

 

So, results in these two cases are almost same.

 

Also, it does not appear that boys and girls have very different birth weights.