Exhibit 9-5

n = 16 H0: μ ≥ 80
= 75.607 Ha: μ

Question

Exhibit 9-5

n = 16	H0: μ ≥ 80
 	 = 75.607	Ha: μ < 80
 	σ = 8.6

Assume the population is normally distributed.

Refer to Exhibit 9-5. The p-value is equal to   _____.

BlackTom AI · Accepted Answer

We start by restating what is given and what is being asked. The exhibit provides: n = 16, x̄ = 75.607, σ = 8.6, H0: μ ≥ 80, Ha: μ < 80, and assumes normal distribution.

Step 1: Compute the test statistic under H0 using the z (or t) approach since σ is given. The standard error is σ/√n = 8.6/√16 = 8.6/4 = 2.15. The observed deviation from the null is x̄ − μ0 = 75.607 − 80 = −4.393.

Step 2: Calculate the z (or t) value. z = (x̄ − μ0) / (σ/√n) = (−4.393) / 2.15 ≈ −2.043. If we used a t distribution (which would be ∼ t15 here since n−1 = 15), the value would be t ≈ −2.043 as well because the standard error uses s ≈ σ when σ is known; the numbers end up very similar.

Step 3: Determine the p-value for Ha: μ < 80 (left-tailed test). For a standard normal, P(Z ≤ −2.043) ≈ 0.0206. For a t15 distribution, P(T15 ≤ −2.043) is in a very similar ballpark (roughly around 0.028, depending on the exact t-cutoff), but because σ is provided and the setup uses a z-method here, the one-sided p-value is approximately 0.0206.

Now, evaluate each answer option against this p-value:
- Option ".0414": This is approximately a two-tailed p-value for |Z| ≈ 2.04, since two-sided p ≈ 2 × 0.0206 ≈ 0.0412. However, the alternative hypothesis is one-sided ( Ha: μ < 80 ), so this two-tailed value is not appropriate for a one-sided test. Therefore this option is not correct for the given one-sided test.
- Option "–.0166": A p-value cannot be negative. This option is nonsensical and clearly incorrect.
- Option "None of the other three": This would be the choice if none of the listed p-values matches the one-sided p-value, or if the available one-sided p-value is not among the provided options.
- Option ".0166": This is a small positive p-value, slightly smaller than the computed ~0.0206. It represents a one-sided p-value that is in the same general vicinity as the computed figure, but the exact calculation with the given numbers yields about 0.0206, not 0.0166. Depending on rounding or if a different distribution (e.g., t with df 15) was used, some slight variation could occur, but 0.0166 is not the precise one-sided p-value from the standard calculation here.

In summary, for a one-sided Ha: μ < 80 with the presented data, the precise p-value is around 0.0206. Among the options, the closest one that matches the one-sided interpretation is not exact, and the two-sided option (0.0414) corresponds to a different testing framework. The negative option is invalid, and if none of the listed values exactly matches the one-sided p-value, some students would pick the choice signaling that none of the other three options is correct.

Note: If the test were interpreted strictly as one-sided with the standard normal reference, the expected p-value would be about 0.0206, which is not exactly any of the listed choices except potentially under rounding or different distribution assumptions.

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Exhibit 9-5 n = 16 H0: μ ≥ 80 = 75.607 Ha: μ < 80 σ = 8.6 Assume the population is normally distributed. Refer to Exhibit 9-5. The p-value is equal to _____.

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