Kindergarten children have heights that are approximately distributed normal. A random sample of size 20 is taken and the mean x and the standard deviation s are calculated ( x = 40 inches and s = 3).
a. Is there sufficient evidence to indicate that the mean of kindergarten children’ height exceeds 40 inches? Perform a test
using alpha=.05.
b. What is the probability that the mean of kindergarten children’ heights would be between 39 and 42.5 inches?
a. In a normal distribution, 50% are above the mean. For confidence interval,
95% = mean ± 1.96 SEm
b. Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability between the two Z scores.
a. To determine whether there is sufficient evidence to indicate that the mean of kindergarten children's height exceeds 40 inches, we can perform a hypothesis test.
First, let's state the null hypothesis (H0) and the alternative hypothesis (Ha):
H0: μ ≤ 40 (the mean height of kindergarten children is less than or equal to 40 inches)
Ha: μ > 40 (the mean height of kindergarten children exceeds 40 inches)
Next, we need to calculate the test statistic, which follows a t-distribution in this case. The formula for the test statistic is:
t = (x - μ) / (s / √n)
where x is the sample mean (40 inches), μ is the population mean (40 inches), s is the sample standard deviation (3 inches), and n is the sample size (20).
Plugging in the values, we get:
t = (40 - 40) / (3 / √20) = 0 / 0.67082 = 0
Since the test statistic is 0, we can conclude that there is not sufficient evidence to indicate that the mean of kindergarten children's height exceeds 40 inches.
b. To find the probability that the mean of kindergarten children's heights would be between 39 and 42.5 inches, we need to calculate the z-scores for these values and use the standard normal distribution.
First, let's calculate the z-scores:
z1 = (39 - 40) / (3 / √20) ≈ -1.8882
z2 = (42.5 - 40) / (3 / √20) ≈ 1.4596
Next, we can use a standard normal distribution table or a calculator to find the probabilities associated with these z-scores.
Using a standard normal distribution table, we can find the probability associated with z1:
P(Z ≤ -1.8882) ≈ 0.0302
Using the same table, we can find the probability associated with z2:
P(Z ≤ 1.4596) ≈ 0.926
To find the probability between these two values, we can subtract the probability associated with z1 from the probability associated with z2:
P(39 ≤ X ≤ 42.5) = P(Z ≤ 1.4596) - P(Z ≤ -1.8882) ≈ 0.926 - 0.0302 ≈ 0.8968
Therefore, the probability that the mean of kindergarten children's heights would be between 39 and 42.5 inches is approximately 0.8968, or 89.68%.