I need help with this these two problem PLEASE!!!
For a population with a mean of μ � 80 and a standard
deviation of � � 12, find the z-score corresponding to
each of the following samples.
a. M � 83 for a sample of n � 4 scores
b. M � 83 for a sample of n � 16 scores
c. M � 83 for a sample of n � 36 scores
A population forms a normal distribution with a mean
of μ � 80 and a standard deviation of � � 15. For
each of the following samples, compute the z-score for
the sample mean and determine whether the sample
mean is a typical, representative value or an extreme
value for a sample of this size.
a. M � 84 for n � 9 scores
b. M � 84 for n � 100 scores
Z = (score-mean)/SEm
SEm = SD/√n
To find the z-score, we will use the formula:
z = (X - μ) / (σ / √n)
where z is the z-score, X is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
Let's calculate the z-scores for each of the given samples:
a. M = 83 for a sample of n = 4 scores
Using the formula: z = (83 - 80) / (12 / √4)
= 3 / (12 / 2)
= 3 / 6
= 0.5
b. M = 83 for a sample of n = 16 scores
Using the formula: z = (83 - 80) / (12 / √16)
= 3 / (12 / 4)
= 3 / 3
= 1
c. M = 83 for a sample of n = 36 scores
Using the formula: z = (83 - 80) / (12 / √36)
= 3 / (12 / 6)
= 3 / 2
= 1.5
Now, let's move on to the next set of problems:
a. M = 84 for n = 9 scores
Using the formula: z = (84 - 80) / (15 / √9)
= 4 / (15 / 3)
= 4 / 5
= 0.8
Since the z-score is within the range of -1.96 to +1.96, the sample mean of 84 is a typical representative value for a sample of size 9.
b. M = 84 for n = 100 scores
Using the formula: z = (84 - 80) / (15 / √100)
= 4 / (15 / 10)
= 4 / 1.5
≈ 2.67
Since the z-score is greater than +1.96, the sample mean of 84 is an extreme value for a sample of size 100.
Remember, the z-score tells us how many standard deviations a sample mean is away from the population mean. Typically, z-scores between -1.96 and +1.96 are considered to be within a typical range. Anything outside this range can be considered extreme.