The true average diameter of ball bearings of a certain type is supposed to be 0
ID: 3310563 • Letter: T
Question
The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out to see whether this is the case. What conclusion is appropriate in each of the following situations? (a) n-15, t-1.64, -0.05 Reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in. o Reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. o Do not reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in. Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. (b) n = 15, t =-1.64, = 0.05 o Reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.S in. Reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. Do not reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in. Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. (c) n = 22, t =-2.4, = 0.01 Reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in. Reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. Do not reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in. o Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. (d) 22,t--3.94 Reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in. o Reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. Do not reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in.Explanation / Answer
a) For n-1 = 14 degrees of freedom, we get the p-value here as:
p = 2P( t14 > 1.64 ) = 2*0.0616 = 0.1232
As the p-value here is 0.1232 > 0.05 which is the level of significance here, therefore the test is insignificant, and we cannot reject the null hypothesis here. Therefore D is the correct answer here.
b) For n-1 = 14 degrees of freedom, we get the p-value here as:
p = 2P( t14 < -1.64 ) = 2*0.0616 = 0.1232
As the p-value here is 0.1232 > 0.05 which is the level of significance here, therefore the test is insignificant, and we cannot reject the null hypothesis here. Therefore D is the correct answer here.
c) For n-1 = 21 degrees of freedom, we get the p-value here as:
p = 2P( t21 < -2.4 ) = 2*0.0129 = 0.0258
As the p-value here is 0.0258 > 0.01 which is the level of significance here, therefore the test is insignificant, and we cannot reject the null hypothesis here. Therefore D is the correct answer here.
d) For n-1 = 21 degrees of freedom, we get the p-value here as:
p = 2P( t21 < -3.94 ) = 2*0.0004 = 0.0008
As the p-value here is 0.0008 < 0.01 which is the level of significance here, therefore the test is significant, and we can reject the null hypothesis here. Therefore A is the correct answer here.
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