Given that: Population mean, μ = 8 Sample mean, x = 7.9 Population standard deviation, o = 0.6 Sample size, n = 33 Level of significance, a = 0.10 Step 1: State the hypotheses: Ho : a = 8 H :μ # 8 This is two-tailed test. Step 2: Compute the test statistic: From CLT, if n >= 30, then sample mean is approximately normally distributed with mean, μ = μ and 0 = 3-4₂ Z= %7 7.9-8 0.6 =-0.957 Step 3: Rejection rule: Reject H0 when p-value 0. 10 Decision: Do not reject HO Step 6: Conclusion: There is not enough evidence to reject the claim at 0.05 significance level.

MATLAB: An Introduction with Applications
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EXPLAIN STEPS 1, 2, AND 3 please do this typewritten and completely so I will upvote. Thank you and much appreciated for the help.

Given that:
Population mean, μ = 8
Sample mean, x = 7.9
Population standard deviation, o = 0.6
Sample size, n = 33
Level of significance, a = 0.10
Step 1:
State the hypotheses:
Ho : a = 8
H₁μ #8
This is two-tailed test.
Step 2:
Compute the test statistic:
From CLT, if n >= 30, then sample mean is approximately normally distributed with mean, μ = μ and
0 =
Step 3:
Rejection rule:
Reject H0 when p-value <a, otherwise do not reject it.
Step 4:
Compute the P-value by using the normal distribution table:
2P (Z ≤ z)=2P (Z ≤ -0.957)
= 2 (0.169)
=0.338
The P-value is 0.338.
Step 5: Make Decision
0.338 > 0. 10
Decision:
Do not reject HO
Step 6:
Conclusion:
There is not enough evidence to reject the claim at 0.05 significance level.
I-H₂
67
7.9-8
0.6
=
√33
= -0.957
Transcribed Image Text:Given that: Population mean, μ = 8 Sample mean, x = 7.9 Population standard deviation, o = 0.6 Sample size, n = 33 Level of significance, a = 0.10 Step 1: State the hypotheses: Ho : a = 8 H₁μ #8 This is two-tailed test. Step 2: Compute the test statistic: From CLT, if n >= 30, then sample mean is approximately normally distributed with mean, μ = μ and 0 = Step 3: Rejection rule: Reject H0 when p-value <a, otherwise do not reject it. Step 4: Compute the P-value by using the normal distribution table: 2P (Z ≤ z)=2P (Z ≤ -0.957) = 2 (0.169) =0.338 The P-value is 0.338. Step 5: Make Decision 0.338 > 0. 10 Decision: Do not reject HO Step 6: Conclusion: There is not enough evidence to reject the claim at 0.05 significance level. I-H₂ 67 7.9-8 0.6 = √33 = -0.957
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