Critical Values for the Student's t Distribution Table to find the critical value or values for the following valuesUseof the significance level o, sample size n, and alternate hypothesis Hi(a) o 0.025, n = 12, H1 >Ho(b) o 0.025, n= 16, HHHo(c) a 0.10. n6. H1 Ho(d) a 0.10. n 28, H1 HHo

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Asked Nov 5, 2019
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Critical Values for the Student's t Distribution Table to find the critical value or values for the following values
Use
of the significance level o, sample size n, and alternate hypothesis Hi
(a) o 0.025, n = 12, H1 >Ho
(b) o 0.025, n= 16, H
HHo
(c) a 0.10. n6. H1 Ho
(d) a 0.10. n 28, H1 HHo
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Critical Values for the Student's t Distribution Table to find the critical value or values for the following values Use of the significance level o, sample size n, and alternate hypothesis Hi (a) o 0.025, n = 12, H1 >Ho (b) o 0.025, n= 16, H HHo (c) a 0.10. n6. H1 Ho (d) a 0.10. n 28, H1 HHo

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Expert Answer

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 (tt0.025) = 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 t0.025­ as 2.20098 ≈ 2.201.

Thus, the critical value is t0.025­ = 2.201.

It is very unlikely to have a test statistic value greater than t0.025­. 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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