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 = 1.69, alpha = 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 = 1.69, alpha = 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. (c) n = 23 t = 2.47, alpha = 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 = 3.94 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 0.5 in.

Explanation / Answer

a) t critical value = 2.201

Here t value < t critical value, So we accept H0

Correct Answer: Option (D)

b)

t critical value = 2.201

Here |t value | < t critical value, So we accept H0

Correct Answer: Option (D)

3)

t critical value = 2.818

Here |t value | < t critical value, So we accept H0

Correct Answer: Option (D)

4)

t critical value = 2.818

Here |t value | > t critical value, So we do not accept H0

Correct Answer: Option (A)

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