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.

Solution :

The null and the alternate hypothesis can be stated as :

Null hypothesis

Alternate hypothesis

We known;

   = 62.286

   = 72.125

   = 116.571

   = 23.45

Therefore, calculating the test statics :

 = -2.312

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

P=T.DIST(t, df)

P=T.DIST(t, df)

 =T.DIST(-2.31, 7)

 = 0.0270

Thus, the of the test is P = 0.0270 is than the level of significance . 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.

square root of 36/6 = 2.449489743

square root of 16/6 = 1.632993162

square root of 6 = 2.449489743

the answer is square root of 16/6 because the square root of 36/6 is the same answer as square root of 6 = 2.449489743

900 degrees

step-by-step explanation: