The true average diameter of ball bearings of a certain type is supposed to be 0

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 = 12, t = 1.63, = 0.05 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. Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. (b) n = 12, t =-1.63, = 0.05 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 5 in. (c) n-23, t =-2.5, = 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 Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in (d) n-23, t=-3.89 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 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 0.5n

Explanation / Answer

Q1) P value = P(t<-1.63)+P(t>1.63)

P value = 0.1314

Option D is Correct ( because p value is greater than 0.05)

Q2) P value = P(t<-1.63)+P(t>1.63)

P value = 0.1314

Option D is Correct ( because p value is greater than 0.05)

Q3) P value = P(t<-2.5)+P(t>2.5)

P value = 0.0204

Option D is Correct ( because p value is greater than 0.01)

Q4) P value = P(t<-3.89)+P(t>3.89)

P value = 0.0008

Option A is Correct ( because p value is very less)

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