An ANOVA procedure is used for data obtained from five populations. Five samples, each comprised of 20 observations, were taken
1 answer:
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Answer:
(a) P(12.99 ≤ X ≤ 13.01) = 0.3840
(b) P(X ≥ 13.01) = 0.3075
Step-by-step explanation:
To address this problem, one needs to grasp the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems involving normally distributed samples are resolved through the application of the z-score formula.
In a data set with a mean and standard deviation
, the z-score for a measurement X is defined as:
The Z-score indicates how many standard deviations the measurement deviates from the mean. After determining the Z-score, one can consult the z-score table to find the relevant p-value for that score. This p-value represents the probability that the measure's value is less than X, which indicates the percentile of X. By subtracting this p-value from 1, we obtain the likelihood that the measure's value exceeds X.
Central Limit Theorem
According to the Central Limit Theorem, for a normally distributed random variable X with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated by a normal distribution, characterized by mean
and standard deviation
.
The Central Limit Theorem can also be applicable to a skewed variable, provided that n is 30 or more.
In this case, we are given that:
(a) Compute P(12.99 ≤ X ≤ 13.01) when n = 16.
In this scenario, we find
This probability equates to the p-value of Z at X = 13.01 minus the p-value of Z at X = 12.99.
X = 13.01
According to the Central Limit Theorem
yields a p-value of 0.3075
0.6915 - 0.3075 = 0.3840
P(12.99 ≤ X ≤ 13.01) = 0.3840
(b) What is the probability that the sample mean diameter is greater than 13.01 when n = 25?
P(X ≥ 13.01) =
This is obtained by subtracting the p-value of Z when X = 13.01 from 1. Hence
has a p-value of 0.6915
1 - 0.6915 = 0.3075
P(X ≥ 13.01) = 0.3075