Chapter 10.2, Problem 10.50E

To determine

To Explain: that the perform a complete analysis of Darwin’s data that includes data a significance test and a confidence interval and follow the four-step statistical.

Answer to Problem 10.50E

There is enough evidence to reject the claim that self-fertilizing and cross-fertilizing leads to the same height in plant

Explanation of Solution

Given:

  

Formula used:

  $\overline{x}=\frac{{\displaystyle \sum x}}{n}$

  $t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}$

  $E=t_{\alpha /2}\times \frac{s}{\sqrt{n}}$

For confidence intervals

  $\begin{array}{c}\overline{x}-E\\ \overline{x}+E\end{array}$

Calculation:

Let us assume:

  $\begin{array}{l}\alpha =0.05\\ c=95\%=0.95\end{array}$

Step 1:

Here are going to perform a significance test the claim that there is no difference in the two groups and it is calculated a $95\%$ confidence interval for the mean difference.

Step 2:

Collecting the data

The sample contains 15 pairs:

  $n=15$

The difference between self-fertilized and cross- fertilized for each

Sample 1	Sample 2	Difference D
23.5	17.4	6.1
12	20.4	-8.4
21	20	3
22	20	2
19.1	18.4	0.7
21.5	18.6	2.9
22.1	18.6	3.5
20.4	15.3	5.1
18.3	16.5	1.8
21.6	18	3.6
23.3	16.3	7
21	18	3
22.1	12.8	9.3
23	15.5	7.5
12	18	-6

The mean is

  $\overline{x}=\frac{{\displaystyle \sum x}}{n}$

  $\begin{array}{c}\overline{x}=\frac{6.1-8.4+1+2+\mathrm{...}+3+9.3+7.5-6}{15}\\ =\frac{39.1}{15}\\ =2.6067\end{array}$

The variance is

  $\begin{array}{c}s=\sqrt{\frac{{\left(6.1-2.6067\right)}^2+\mathrm{...}+{\left(-6-2.6067\right)}^2}{15-1}}\\ =4.7128\end{array}$

Step 3:

Analyzing the data

SIGNIFICANCE TEST

The claim is either the null hypothesis or the alternative hypothesis .the null hypothesis statement is that the population mean is equal to the value given in the claim. If the null hypothesis is the claim, then the alternative hypothesis statement is the opposite of null hypothesis.

  $\begin{array}{c}{H}_0:\mu =0\\ H_1:\mu >0\end{array}$

The value of the test statistic is

  $\begin{array}{c}t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}\\ =\frac{2.6067-0}{4.7128/\sqrt{15}}\\ =2.142\end{array}$

The P-value is the probability of getting the value of the test statistic, or a value more extreme; assume that the null hypothesis is true. For the P-value $df=n-1=15-1=14:$

  $0.025<P<0.05$

If the P-value is lesser than the significance level $\alpha$ , then the alternative hypothesis is accepted and null hypothesis is rejected.

  $P<0.05\Rightarrow {\text{Reject H}}_0$

The t-value by looking in the row starting with degrees of freedom $df=n-1=15-1=14$ and in the column with $p=\left(1-c\right)/2=0.025$ in the table of the $T$ distribution:

  $t_{\alpha /2}=2.145$

The margin of error is

  $\begin{array}{c}E=t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =2.145\times \frac{4.7128}{\sqrt{15}}\\ =2.6101\end{array}$

The confidence intervals are

  $\begin{array}{c}\overline{x}-E=2.6067-2.6101=-0.0034\\ \overline{x}+E=2.6067+2.6101=5.2168\end{array}$

Step 4:

There is enough evidence to reject the claim that self-fertilizing and cross-fertilizing leads to the same height in plant. There are 95% confident that the mean height of the cross-fertilization is between $0.0034$ inches lower and $5.2168$ inches higher than the mean height of self-fertilization.