In independent golf equipment testing facility compared the difference in the performance of golf balls hit off a regular \(2-3/4\)" wooden tee to those hit off a \(3\)" Stringer Competition golf tee. A Callaway Great Big Bertha driver with \(10\) degrees of loft was used for the test, and a robot swung the club head at approximately \(95\) miles per hour. Data on total distance travelled (in yards) with each type of tee, based on the test results are provided on the WeissStats site.

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

b. At the \(1%\) significance level, do the data provide sufficient evidence to conclude that on average the Stringer tee improves total distance travelled?

c. Find \(99%\) confidence interval for the difference between the mean total distance travelled with the regular and Stringer tees.

d. Are your procedures in part (b) and (c) justified? Why or why not?

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

vegetarian females \(=30\)

omnivores females \(=30\)

mean \(=[227.2, 240.9333]\)

Standard deviation \(=2.1399 2.7784]\)

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

\(x_{1}=norminv(rand(30,1),227,2.1)\)

\(x_{2}=norminv(rand(30,1),241,2.7)\)

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.

The null hypothesis needs to contain an equality

\(H_{0}:\mu_{1}=\mu_{2}\)

\(H_{a}:\mu_{1}<\mu_{2}\)

Calculate the pooled standard deviation

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

Put all the values and solve

\(s_{p}=\sqrt{\frac{(30-1)(2.1399)^{2}+(30-1)(2.7784)^{2}}{30+30-2}}\)

\(S_{p}=2.4798\)

Determine the test statistics

\(t=\frac{\bar{x_{1}}-\bar{x_{2}}}{s_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}=\frac{227.2-240.9333}{2.4798\sqrt{\frac{1}{30}+\frac{1}{30}}}\approx -21.449\)

The degree of freedom will be

\(df=n_{1}+n_{2}-2=30+30-2=58\)

So, the \(P-\)value will be

\(P<0.005\)

If the p-value is less than the significance level then the null hypothesis will be rejected

\(P<0.01\Rightarrow\) Reject \(H_{0}\)

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

\(s_{p}=\sqrt{\frac{(30-1)(2.1399)^{2}+(30-1)(2.7784)^{2}}{30+30-2}}\)

\(S_{p}=2.4798\)

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.