Respuesta :
Solution :
The null and the alternate hypothesis can be stated as :
Null hypothesis
[tex]$H_0:\mu_1 \geq \mu_2$[/tex]
Alternate hypothesis
[tex]$H_a:\mu_1 \leq \mu_2$[/tex]
We known;
[tex]$\overline x_1=\frac{\sum_{i=1}^n X_i}{n_1}$[/tex]
[tex]$=\frac{43+....+75}{7}$[/tex]
= 62.286
[tex]$\overline x_2=\frac{\sum_{i=1}^n X_i}{n_2}$[/tex]
[tex]$=\frac{66+....+81}{16}$[/tex]
= 72.125
[tex]$s_1^2=\frac{\sum_{i=1}^n(X_i- \overline X_1)^2}{n_1-1}$[/tex]
[tex]$=\frac{(43-65.5)^2+....+(75-65.5)^2}{7-1}$[/tex]
= 116.571
[tex]$s_2^2=\frac{\sum_{i=1}^n(X_i- \overline X_2)^2}{n_2-1}$[/tex]
[tex]$=\frac{(66-72.13)^2+....+(81-72.13)^2}{16-1}$[/tex]
= 23.45
Therefore, calculating the test statics :
[tex]$t=\frac{\overline x_1 - \overline x_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$[/tex]
[tex]$t=\frac{62.29-72.125}{\sqrt{\frac{116.571}{7}+\frac{23.45}{16}}}$[/tex]
[tex]$t=\frac{-9.839}{4.2566}$[/tex]
= -2.312
Now calculating the P-value for the test as follows :
P=T.DIST(t, df)
[tex]$df=\frac{\left(\frac{s_1^2}{n_1}+\frac{s^2_2}{n_2}\right)^2}{\frac{1}{n_1-1}\left(\frac{s^2_X}{n_1}\right)^2+\frac{1}{n_2-1}\left(\frac{s^2_Y}{n_2}\right)^2}$[/tex]
[tex]$df=\frac{\left(\frac{116.571}{7}+\frac{23.45}{16}\right)^2}{\frac{1}{7-1}\left(\frac{116.571}{7}\right)^2+\frac{1}{16-1}\left(\frac{23.45}{16}\right)^2}$[/tex]
[tex]$=\frac{328.2868}{46.36395}$[/tex]
[tex]$\approx 7$[/tex]
P=T.DIST(t, df)
=T.DIST(-2.31, 7)
= 0.0270
Thus, the [tex]$\text{P-value}$[/tex] of the test is P = 0.0270 is [tex]$\text{less}$[/tex] than the level of significance [tex]$\alpha= 0.05$[/tex]. Hence the researcher can reject the null hypothesis.
Conclusion: The mean launch temperature for the flights with O ring damages less than that for the flights with no O rings.
