Question

Asked Nov 5, 2019

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Step 1

**(a)**

The *t*-test considered here is a one-tailed test.

The sample size is, *n* = 12.

Since one of the degrees of freedom is used up for estimating the true population mean (by finding the sample mean), the degrees of freedom for this test is, *n* – 1 = 12 – 1 = 11.

The level of significance is, *α* = 0.025.

Since this is a one-tailed test, there must be one critical value.

The critical values are such that, *P* (*t* ≥ *t*_{0.025}) = 0.025.

Step 2

**Critical value calculation and rejection region:**

The Excel formula for the cumulative probability in this case, **=T.INV(0.975,11) **gives the value of *t*_{0.025} as 2.20098 ≈ 2.201.

Thus, the critical value is *t*_{0.025} = **2.201**.

It is very unlikely to have a test statistic value greater than *t*_{0.025}. Thus, the rejection region (for rejecting the relevant null hypothesis) is ** t > 2.201**.

Step 3

**(b)**

The *t*-test considered here is a one-tailed test.

The sample size is, *n* = 16.

Since one of the degrees of freedom is used up for estimating the true population mean (by finding the sample mean), the degrees of freedom for this test is, *n* – 1 = 16 – 1 = 15.

The le...

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