Question

Space shuttle Challengerexploded because of O-ring failure shortly after it was launched. O-ring damage and temperature at time of launch for the 23 space shuttle flights that preceded the Challenger. The data is reproduced below.

Flights with O-ring damage 43 57 58 63 70 70 75
Flights with no O-ring damage 66 67 67 67 68 69 70 70 72 73 75 76 76 78 79 81
Is the mean launch temperature for flights with O-ring damage significantly less than for flights with no O-ring damage? Use 5% level of significance.

Answer 1

The null and the alternate hypothesis can be stated as :

Null hypothesis

`$H_0:\mu_1 \geq \mu_2$`

Alternate hypothesis

`$H_a:\mu_1 \leq \mu_2$`

We known;

`$\overline x_1=(\sum_(i=1)^n X_i)/(n_1)$`

`$=(43+....+75)/(7)$`

= 62.286

`$\overline x_2=(\sum_(i=1)^n X_i)/(n_2)$`

`$=(66+....+81)/(16)$`

= 72.125

`$s_1^2=(\sum_(i=1)^n(X_i- \overline X_1)^2)/(n_1-1)$`

`$=((43-65.5)^2+....+(75-65.5)^2)/(7-1)$`

= 116.571

`$s_2^2=(\sum_(i=1)^n(X_i- \overline X_2)^2)/(n_2-1)$`

`$=((66-72.13)^2+....+(81-72.13)^2)/(16-1)$`

= 23.45

Therefore, calculating the test statics :

`$t=\frac{\overline x_1 - \overline x_2}{\sqrt{(s_1^2)/(n_1)+(s_2^2)/(n_2)}}$`

`$t=\frac{62.29-72.125}{\sqrt{(116.571)/(7)+(23.45)/(16)}}$`

`$t=(-9.839)/(4.2566)$`

= -2.312

Now calculating the P-value for the test as follows :

P=T.DIST(t, df)

`$df=(\left((s_1^2)/(n_1)+(s^2_2)/(n_2)\right)^2)/((1)/(n_1-1)\left((s^2_X)/(n_1)\right)^2+(1)/(n_2-1)\left((s^2_Y)/(n_2)\right)^2)$`

`$df=(\left((116.571)/(7)+(23.45)/(16)\right)^2)/((1)/(7-1)\left((116.571)/(7)\right)^2+(1)/(16-1)\left((23.45)/(16)\right)^2)$`

`$=(328.2868)/(46.36395)$`

`$\approx 7$`

P=T.DIST(t, df)

=T.DIST(-2.31, 7)

= 0.0270

Thus, the
`$\text{P-value}$` of the test is P = 0.0270 is
`$\text{less}$` than the level of significance
`$\alpha= 0.05$`. 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.