Philosophical and health issues are prompting an increasing number of Taiwanese to switch to a vegetarian lifestyle. In the paper "LDL of Taiwanese Vegetarians Are Less Oxidizable than Those of Omnivores" (Journal of Nutrition, Vol. 130, Pp. 1591-1596), S. Lu et al. compared the daily intake of nutrients by vegetarians and omnivores living in Taiwan. Among the nutrients considered was protein. Too little protein stunts growth and interferes with all bodily functions; too much protein puts a strain on the kidneys, can cause diarrhea and dehydration, and can leach calcium from bones and teeth. Independent random samples of \(51\) female vegetarians and \(53\) female omnivores yielded the data, in grams, (n daily protein intake presented on the WeissStats site. Use the technology of your choice to do the following.

a. Obtain normal probability plots, boxplots, and the standard deviations for the two samples.

b. Do the data provide sufficient evidence to conclude that the mean daily protein intakes of female vegetarians and female omnivores differ? Perform the required hypothesis test at the \(1%\)significance level.

c. Find a \(99%\) confidence interval for the difference between the mean daily protein intakes of female vegetarians and female omnivores.

d. Are your procedures in parts (b) and (c) justified? Explain your answer.

The number of sample \((n)\), mean \((\bar{x})\) and standard deviation \((\sigma)\) is given.

vegetarian females \(=51\)

omnivores females \(=53\)

mean \(=[39, 50]\)

Standard deviation \(=18 19]\)

Generate \(51\) and \(53\) samples of vegetarian and omnivores females using function “norminv” with given mean and standard deviation in MATLAB

\(x_{1}=norminv(rand(51,1),39,18)\)

\(x_{2}=norminv(rand(51,1),50,19)\)

After that we will get random samples and generate the normal probability and boxplot.

Program:

Query:

• First, we have defined the number of samples.
• Then create random samples using function “norminv” with given mean and standard deviation.

• Sketch a boxplot and normal probability plot.

Calculate the mean for vegetarian females using relation

\(\bar{x_{1}}=\frac{\sum_{i-1}^{n}x_{i}}{n}\)

Put all the values and solve

\(\bar{x_{1}}=\frac{\sum_{i-1}^{n}x_{i}}{n}=\frac{39+57+...+28+46}{51}\)

After solving we will get

\(=39.0392\)

Similarly, for omnivores females

\(\bar{x_{2}}=\frac{\sum_{i-1}^{n}x_{i}}{n}=\frac{55+39+...+57+5}{53}=49.9245\)

Calculate the standard deviation for both the females using relation

\(s_{1}=\sqrt{\frac{\sum_{i-1}^{n}(x_{i}-\bar{x_{1}})^{2}}{n-1}}=39.0392\)

\(s_{2}=\sqrt{\frac{\sum_{i-1}^{n}(x_{i}-\bar{x_{1}})^{2}}{n-1}}=49.9245\)

The mean is

\(\bar{x_{1}}=\frac{\sum_{i-1}^{n}x_{i}}{n}=\frac{39+57+...+28+46}{51}=39.0392\)

\(\bar{x_{2}}=\frac{\sum_{i-1}^{n}x_{i}}{n}=\frac{55+39+...+57+5}{53}=49.9245\)

And the standard deviation is

\(s_{1}=\sqrt{\frac{\sum_{i-1}^{n}(x_{i}-\bar{x_{1}})^{2}}{n-1}}=39.0392\)

\(s_{2}=\sqrt{\frac{\sum_{i-1}^{n}(x_{i}-\bar{x_{1}})^{2}}{n-1}}=49.9245\)

To calculate the \(99%\) confidence interval, we will calculate the degree of freedom

\(df=n_{1}+n_{2}-2=51+53-2=102\)

The student t distribution will be

\(t_{\alpha /2}=2.626\)

Determine the pool standard deviation using relation

\(s_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}}\)

Put the all the values of resolve the relation

\(s_{p}=\sqrt{\frac{(51-1)(18.8218)^{2}+(53-1)(18.9684)^{2}}{n_{1}+n_{2}-2}}\)

\(S_{p}=18.8967\)

Calculate the marginal error using relation

\(E=t_{\alpha /2}s\sqrt {\frac{1}{n_{1}}+\frac{1}{n_{2}}}\)

After solving we will get

\(E=2.626\times 181.8967\sqrt {\frac{1}{51}+\frac{1}{53}}\)

\(E\approx 9.7336\)

Calculate the 99% confidence interval

\((\bar{x_{1}}-\bar{x_{2}})-E=(39.0392-49.9245)-9.7336=20.6189\)

\((\bar{x_{1}}-\bar{x_{2}})+E=(39.0392-49.9245)+9.7336=-1.1517\)

The 99% confidence interval will be

\(99%\) confidence interval is \([20.6189, -1.1517]\)

If the normal probability plot is linear, then the data come from a population with a normal distribution.