from a normally distributed infinite number of iron bars lengths with a mean and standard deviation as 4 ft. and 0.6 ft. respectively, a sample of 100 bars is taken. if the sample mean is 4.2 ft., can the sample be called a truely random sample?
How was the sample taken? If the observations are taken on a chance basis, then it would be random.
Or are you asking if the sample is within certain range of possible sample means?
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.
To determine if the sample can be called a truly random sample, we need to assess if the sample mean of 4.2 ft. is significantly different from the population mean of 4 ft., considering the standard deviation of 0.6 ft.
We can conduct a hypothesis test to evaluate this scenario. Let's assume the null hypothesis (H0) is that the sample mean is equal to the population mean, and the alternative hypothesis (Ha) is that the sample mean is not equal to the population mean.
H0: 𝜇 (Sample Mean) = 4 ft.
Ha: 𝜇 (Sample Mean) ≠ 4 ft.
We can use the z-test statistic to determine the likelihood of obtaining the sample mean of 4.2 ft. if the null hypothesis is true. The formula for the z-test statistic is as follows:
z = (sample mean - population mean) / (standard deviation / square root of sample size)
z = (4.2 - 4) / (0.6 / sqrt(100))
z = 0.2 / (0.6 / 10)
z = 0.2 / 0.06
z = 3.33
Using a z-table or statistical software, we can determine the critical z-value for a significance level (α) of 0.05 (assuming a two-tailed test). The critical z-value is approximately ±1.96.
Since the calculated z-value of 3.33 is greater than the critical z-value of 1.96, we can reject the null hypothesis. This means that the sample mean of 4.2 ft. is significantly different from the population mean of 4 ft.
Therefore, based on the sample mean of 4.2 ft., we can conclude that the sample is likely not a truly random sample and may not represent the population accurately.
To determine whether the sample can be called a truly random sample, we need to conduct a hypothesis test. In this case, we can use a one-sample t-test.
First, we need to define our null hypothesis (H0) and alternative hypothesis (Ha):
- H0: The sample mean is equal to the population mean (μ = 4 ft.)
- Ha: The sample mean is not equal to the population mean (μ ≠ 4 ft.)
Then, we calculate the test statistic using the formula:
t = (sample mean - population mean) / (sample standard deviation / √sample size)
Here are the steps to perform the t-test:
1. Calculate the standard error of the sample mean:
standard error = population standard deviation / √sample size
2. Calculate the t-value:
t = (sample mean - population mean) / standard error
3. Determine the degrees of freedom (df):
df = sample size - 1
4. Look up the critical t-value for the desired significance level and df.
For example, at a 95% confidence level with df = 99, the critical t-value is approximately ±1.984.
5. Compare the calculated t-value with the critical t-value.
- If the calculated t-value falls within the range of the critical t-value, we fail to reject the null hypothesis, meaning the sample can be considered a random sample.
- If the calculated t-value falls outside the range of the critical t-value, we reject the null hypothesis, indicating that the sample is not a random sample.
Now, let's calculate the t-value and determine if the sample can be called a truly random sample:
Given:
- Population mean (μ) = 4 ft.
- Population standard deviation (σ) = 0.6 ft.
- Sample size (n) = 100
- Sample mean (x̄) = 4.2 ft.
1. Calculate the standard error:
standard error = σ / √n
standard error = 0.6 ft. / √100 = 0.06 ft.
2. Calculate the t-value:
t = (x̄ - μ) / standard error
t = (4.2 ft. - 4 ft.) / 0.06 ft.
t = 0.2 ft. / 0.06 ft.
t ≈ 3.33
3. Determine the degrees of freedom (df):
df = n - 1
df = 100 - 1 = 99
4. Look up the critical t-value:
For a 95% confidence level and df = 99, the critical t-value is approximately ±1.984.
5. Compare the calculated t-value with the critical t-value:
|t| > critical t-value (3.33 > 1.984)
Since the calculated t-value (3.33) is greater than the critical t-value (1.984), we reject the null hypothesis. Therefore, we can conclude that the sample is not a truly random sample.