Solved Name: Hypothesis Testing Worksheet MTH 163 Basic | Chegg.com - Free Printable

Let’s solve each part step by step. We’re finding P-values for different hypothesis tests using the standard normal (z) distribution. Remember:

- For a left-tailed test, P-value = area to the left of z*.
- For a right-tailed test, P-value = area to the right of z* = 1 - area to the left of z*.
- For a two-tailed test, P-value = 2 × (area in the tail beyond |z*|). If z* is negative, find area to the left; if positive, find area to the right — then double it.

We’ll use a standard normal table or calculator to find areas. I’ll show you how to think about it.

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(a) Left-tailed test, z* = -1.55

Look up z = -1.55 in the z-table → area to the left is 0.0606.

✔ P-value = 0.0606

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(b) Two-tailed test, z* = 2.63

First, find area to the right of 2.63 → 1 - 0.9957 = 0.0043
Then double it for two tails → 2 × 0.0043 = 0.0086

✔ P-value = 0.0086

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(c) Two-tailed test, z* = -2.63

Same as (b), because we take absolute value for two-tailed. Area to the left of -2.63 is 0.0043 → double it → 0.0086

✔ P-value = 0.0086

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(d) Right-tailed test, z* = 0.78

Area to the left of 0.78 is 0.7823 → so area to the right is 1 - 0.7823 = 0.2177

✔ P-value = 0.2177

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(e) Right-tailed test, z* = 1.85

Area to the left of 1.85 is 0.9678 → area to the right = 1 - 0.9678 = 0.0322

✔ P-value = 0.0322

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(f) Two-tailed test, z* = -0.99

Area to the left of -0.99 is 0.1611 → since it’s two-tailed, we double the smaller tail → 2 × 0.1611 = 0.3222

Wait — actually, for two-tailed, we always double the tail area beyond |z|. So |z| = 0.99 → area to the left of -0.99 is 0.1611, which is the same as area to the right of +0.99. So yes, 2 × 0.1611 = 0.3222.

✔ P-value = 0.3222

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Now let’s summarize all P-values:

(a) 0.0606
(b) 0.0086
(c) 0.0086
(d) 0.2177
(e) 0.0322
(f) 0.3222

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Question (2): Reject or fail to reject at α = 0.05?

Rule: If P-value ≤ α → reject H₀
If P-value > α → fail to reject H₀

Compare each P-value to 0.05:

(a) 0.0606 > 0.05 → fail to reject
(b) 0.0086 < 0.05 → reject
(c) 0.0086 < 0.05 → reject
(d) 0.2177 > 0.05 → fail to reject
(e) 0.0322 < 0.05 → reject
(f) 0.3222 > 0.05 → fail to reject

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Question (3): Reject or fail to reject at α = 0.01?

Compare each P-value to 0.01:

(a) 0.0606 > 0.01 → fail to reject
(b) 0.0086 < 0.01 → reject
(c) 0.0086 < 0.01 → reject
(d) 0.2177 > 0.01 → fail to reject
(e) 0.0322 > 0.01 → fail to reject
(f) 0.3222 > 0.01 → fail to reject

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Final Answer:

(1) P-values:
(a) 0.0606
(b) 0.0086
(c) 0.0086
(d) 0.2177
(e) 0.0322
(f) 0.3222

(2) At α = 0.05:
Reject: (b), (c), (e)
Fail to reject: (a), (d), (f)

(3) At α = 0.01:
Reject: (b), (c)
Fail to reject: (a), (d), (e), (f)