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?
http://davidmlane.com/hyperstat/z_table.html
To answer these questions, we need to perform hypothesis testing and use the properties of the normal distribution.
a. Testing if the mean height exceeds 40 inches:
Step 1: Define the null and alternative hypotheses:
Null hypothesis (H0): The mean height of kindergarten children is equal to or less than 40 inches.
Alternative hypothesis (Ha): The mean height of kindergarten children exceeds 40 inches.
Step 2: Determine the significance level (alpha):
The significance level (alpha) is given as 0.05 or 5%. This means that we are willing to accept a 5% chance of making a Type I error (rejecting the null hypothesis when it is true).
Step 3: Calculate the test statistic:
We will use the t-distribution since the population standard deviation is unknown. The test statistic formula is:
t = (x - μ) / (s / √n)
Where:
x = sample mean
μ = population mean (null hypothesis value)
s = sample standard deviation
n = sample size
Given:
x = 40 inches
s = 3 inches
n = 20
Substituting the values into the formula:
t = (40 - 40) / (3 / √20)
t = 0 / 0.6708
t = 0
Step 4: Determine the critical value or p-value:
Since our alternative hypothesis is one-tailed (exceeds 40 inches), we will find the critical value corresponding to the 95% confidence level (alpha = 0.05) in the upper tail of the t-distribution. The critical value for a one-tailed test with 19 degrees of freedom is approximately 1.729.
Step 5: Make a decision:
If the calculated test statistic falls to the right of the critical value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
Since our test statistic, t = 0, falls within the range (-∞, 1.729), we fail to reject the null hypothesis.
Therefore, there is not enough evidence to indicate that the mean of kindergarten children's height exceeds 40 inches based on the given sample.
b. Finding the probability of the mean height between 39 and 42.5 inches:
Step 1: Standardize the values:
To find the probability, we need to standardize the values using the standard normal distribution.
Z1 = (39 - μ) / (s / √n)
Z2 = (42.5 - μ) / (s / √n)
Step 2: Calculate the z-scores:
Using the given values and the distribution properties:
Z1 = (39 - 40) / (3 / √20) = -0.5747
Z2 = (42.5 - 40) / (3 / √20) = 1.8257
Step 3: Find the cumulative probabilities:
Using the z-scores, we can find the cumulative probabilities using a standard normal distribution table or a statistical calculator.
P(39 < x < 42.5) = P(Z1 < Z < Z2) = P(Z < 1.8257) - P(Z < -0.5747)
Using a standard normal distribution table or a statistical calculator, we can find the probabilities. Let's assume P(Z < 1.8257) = 0.9656 and P(Z < -0.5747) = 0.2838.
P(39 < x < 42.5) = 0.9656 - 0.2838 = 0.6818
Therefore, the probability that the mean height of kindergarten children would be between 39 and 42.5 inches is approximately 0.6818 (or 68.18%).