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Questionome - Northern Ess... Content ww-awu.aleks.com AAleks A_{\text {Aleks }} Blackboard Homework \# 5: 9(1,3,4,5) 14(1,2) Question 5 of 30 (1 point) I Question Attempt: 1 of 3 ✓1\checkmark 1 ×2\times 2 ✓3\checkmark 3 ✓4\checkmark 4 =5=5 6 7 8 9 10
Find the PP-value for the following values of the test statistic tt, sample size nn, and alternate hypothesis H1H_{1}. Use the (3) Critical Values for the Student's t Distribution Table and specify that PP is between two values.
Part 1 of 2 (a) t=−1.498,n=21t=-1.498, n=21, and H1:μ≠μ0H_{1}: \mu \neq \mu_{0} 0.10<P-value <0.200.10<P \text {-value }<0.20
Alternate Answer: 0.1<P-value <0.20.1<P \text {-value }<0.2
Part: 1/21 / 2
Part 2 of 2 (b) t=−1.585,n=15t=-1.585, n=15, and H1:μ<μ0H_{1}: \mu<\mu_{0} 0.10⊗<P-value <0.15⊗0.10^{\otimes}<P \text {-value }<0.15{ }^{\otimes}
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Studdy Solution

STEP 1

1. We are using the Student's t-distribution to find the P-value.
2. The test statistic t t and sample size n n are given.
3. The alternate hypothesis H1 H_1 is specified for each part.
4. We will use a t-distribution table to find the range for the P-value.

STEP 2

1. Identify the degrees of freedom.
2. Use the t-distribution table to find the P-value range for part (a).
3. Use the t-distribution table to find the P-value range for part (b).

STEP 3

Calculate the degrees of freedom for part (a) and part (b).
For part (a), n=21 n = 21 , so degrees of freedom df=n−1=20 df = n - 1 = 20 .
For part (b), n=15 n = 15 , so degrees of freedom df=n−1=14 df = n - 1 = 14 .

STEP 4

For part (a), t=−1.498 t = -1.498 , df=20 df = 20 , and H1:μ≠μ0 H_1: \mu \neq \mu_0 .
Look up the critical values for t=−1.498 t = -1.498 with df=20 df = 20 in the t-distribution table.
The P-value is between 0.10 and 0.20.

STEP 5

For part (b), t=−1.585 t = -1.585 , df=14 df = 14 , and H1:μ<μ0 H_1: \mu < \mu_0 .
Look up the critical values for t=−1.585 t = -1.585 with df=14 df = 14 in the t-distribution table.
The P-value is between 0.10 and 0.15.
The P-value ranges are: - Part (a): 0.10<P-value<0.20 0.10 < P \text{-value} < 0.20 - Part (b): 0.10<P-value<0.15 0.10 < P \text{-value} < 0.15

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