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Minitab Problem 3 Imagine choosing n = 16 women at random from a large population and measuring their heights. Assume that th
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Session Variable C1 C2 C3 C4 C5 C6 C7 Се c9 C10 C11 C12 C13 C14 C15 C16 C17 C18 C19 C20 N Mean St Dev 16 63.278 2.155 16 63.4(a) Answer: 19 (Because we see that P-value of 19 out of 20 tests>0.10 and data corresponding C5 does not support the claim and its P-value=0.072<0.10).

(b) Answer 1 ( the data corresponding C5 does not support the claim and its P-value=0.072<0.10).

(c)

Variable   P-value
C1 0.336
C2 0.430
C3 0.440
C4    0.583
C5    0.072
C6 0.522
C7    0.417
C8 0.385
C9 0.445
C10 0.964
C11    0.196
C12 0.450
C13 0.831
C14 0.235
C15 0.534
C16 0.488
C17 0.493
C18 0.942
C19 0.496
C20 0.555

We see that P-values are not same.

(d) Since P-value for C5>0.05, all P-value>0.05, hence at \alpha =0.05, we fail to reject all 20 tests. However for \alpha =0.10, we fail to reject 19 out of 20 tests.

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