You measure 64 children , obtaining a X of 57.28 . The mean is 56 and sd is 8 Slug says that bc this X is so close to the mean of 56 , this sample could hardly be considered gifted. A. Perform the appropriate statistical procedure to determine whether he is correct
B. In what percentage of the top scores is this sample mean?
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 related to the Z score. Relate to level of significance you are using.
Multiply probability by 100 to get percentage.
A. To determine if Slug is correct, we need to perform a statistical procedure. One common procedure is to conduct a hypothesis test. We can use a t-test to compare the sample mean to the population mean and see if they are significantly different.
Before conducting the t-test, we need to check the assumptions. One assumption is that the variable follows a normal distribution. We also need to check if the sample size is large enough for the t-test to be valid. Generally, a sample size of 30 or more is considered large enough.
Let's assume that the variable follows a normal distribution and the sample size is large enough.
The null hypothesis (H0) for the t-test is that the sample mean is equal to the population mean. The alternative hypothesis (H1) is that the sample mean is not equal to the population mean.
We can perform a two-sample t-test using the formula:
t = (X - μ) / (s / √n)
Where:
X is the sample mean (57.28),
μ is the population mean (56),
s is the standard deviation of the population (8),
n is the sample size (64).
Calculating the t-value:
t = (57.28 - 56) / (8 / √64)
t = 1.28 / (8 / 8)
t = 1.28
The degrees of freedom (df) for the t-test are (n - 1), so df = 63.
We can then find the critical value for the t-test using a t-distribution table or software. Let's assume a significance level of 0.05 (95% confidence level) and a two-tailed test.
Looking up the critical value for df = 63 and a significance level of 0.05, we find a critical value of approximately ±2.00.
Since the calculated t-value (1.28) is less than the critical value (2.00), we fail to reject the null hypothesis.
Therefore, there is not enough evidence to conclude that the sample mean is significantly different from the population mean of 56. Slug's claim that the sample could hardly be considered gifted based on X being close to the mean is not supported by this statistical test.
B. To determine the percentage of top scores that this sample mean represents, we need to calculate the z-score for the sample mean and then find the corresponding percentile from the standard normal distribution.
The formula for calculating the z-score is:
z = (X - μ) / σ
Where:
X is the sample mean (57.28),
μ is the population mean (56),
σ is the standard deviation of the population (8).
Calculating the z-score:
z = (57.28 - 56) / 8
z = 1.28 / 8
z = 0.16
To find the corresponding percentile, we can use a standard normal distribution table or software.
Using a standard normal distribution table, we find that a z-score of 0.16 corresponds to approximately the 56th percentile.
Therefore, this sample mean of 57.28 is in the top 56% of scores.
To determine whether the sample mean of 57.28 is significantly different from the population mean of 56, we can perform a hypothesis test using a t-test.
A. Performing a t-test:
1. Set up the null and alternative hypotheses:
- Null hypothesis (H0): The sample mean (X̄) is equal to the population mean (μ).
- Alternative hypothesis (Ha): The sample mean (X̄) is not equal to the population mean (μ).
2. Set the significance level (α): In this case, let's assume a common significance level of 0.05 (5%).
3. Calculate the test statistic:
- First, calculate the standard error of the mean (SE):
SE = standard deviation (SD) / square root of the sample size (n)
SE = 8 / square root of 64
SE = 1
- Next, calculate the t-value:
t = (X̄ - μ) / SE
t = (57.28 - 56) / 1
t = 1.28
4. Determine the critical value: Look up the critical t-value for a two-tailed test with a 0.05 significance level and degrees of freedom (df) equal to the sample size minus 1 (n - 1). In this case, with n = 64, the critical t-value is approximately ±1.997.
5. Compare the calculated t-value with the critical t-value: If the calculated t-value falls within the critical region (outside of the range defined by the critical values), we reject the null hypothesis. If it falls within the non-critical region, we fail to reject the null hypothesis.
In this case, the calculated t-value of 1.28 does not fall within the critical region. Therefore, we fail to reject the null hypothesis.
B. To determine what percentage of top scores the sample mean of 57.28 represents, we can use a z-score and the standard normal distribution table.
1. Calculate the z-score:
z = (X - μ) / SD
z = (57.28 - 56) / 8
z = 0.285
2. Look up the percentage corresponding to the z-score in the standard normal distribution table. A z-score of 0.285 corresponds to approximately 61.46%.
Therefore, the sample mean of 57.28 is in the top 61.46% of scores.