Answer :
The calculated test statistics (t) is around -0.83, and the comparing p-value is around 0.8972, demonstrating no significant difference. The right reply for the multiple-choice question is p-value > 0.10.
How to find the p-value
a. The competing speculations that decide whether there's any distinction between the normal crop yields from the utilization of diverse fertilizers are:
H0: μD = 0; HA: μD ≠
b. To calculate the esteem of the test statistic, we have to discover the cruel distinction (¯D) and the standard deviation of the contrasts (SD).
Cruel contrast (¯D):
¯D = Σ(D) / n
where D speaks to the contrasts (Ancient - Unused) and n is the number of perceptions.
Calculating the mean difference:
11 - 12 + 11 - 9 + 9 - 13 + 8 - 9 + 13 - 12 + 11 - 12 = -3
(¯D) = (-3 / 6)= -0.5
The standard deviation of the contrasts (SD):
sD = √[Σ(D - ¯D)² / (n - 1)]
where Σ(D - ¯D)² speaks to the whole of squared contrasts.
Calculating the standard deviation of the contrasts:
-0.5 - (-0.5)² + 1.5 - (-0.5)² + -4.5 - (-0.5)² + -1.5 - (-0.5)² + 0.5 - (-0.5)² + -0.5 - (-0.5)² = 22.5
sD = √(22.5 / 5) ≈ 1.89
The test measurement (t) is calculated as:
t = ¯D / (SD / √n)
Substituting the values:
t = (-0.5) / (1.89 / √6) ≈ -0.83
c. With 5 degrees of opportunity (n - 1 = 6 - 1 = 5) and a two-tailed test, the p-value is roughly 0.4486. Since it's a two-tailed test, we twofold this esteem to urge the ultimate p-value.
Hence, the p-value is roughly 0.8972.
The right reply for the multiple-choice question is p-value > 0.10.
Learn more about p-value here:
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