We would like to test at a level of significance α = 0.01 whether the population mean is less than 543. A random sample of 30 has a mean of 540. Assume σ = 10 and normally distributed population. Find the p-value.
Question
We would like to test at a level of significance α = 0.01 whether the population mean is less than 543. A random sample of 30 has a mean of 540. Assume σ = 10 and normally distributed population. Find the p-value.
Solution 1
To find the p-value, we first need to calculate the z-score. The z-score is calculated using the formula:
Z = (X - μ) / (σ / √n)
where: X = sample mean = 540 μ = population mean = 543 σ = standard deviation = 10 n = sample size = 30
Substituting the values into the formula, we get:
Z = (540 - 543) / (10 / √30) = -1.732
Since we are testing whether the population mean is less than 543, we are dealing with a one-tailed test. The p-value is the probability that we would observe a z-score as extreme as -1.732 or more so, assuming the null hypothesis is true.
Looking up the z-score of -1.732 in the z-table, we find that the probability associated with this z-score is 0.0418. However, this is the probability for a two-tailed test. Since we are conducting a one-tailed test, the p-value is simply 0.0418.
Therefore, the p-value is 0.0418. Since the p-value is greater than the level of significance α = 0.01, we fail to reject the null hypothesis. This means that we do not have enough evidence to conclude that the population mean is less than 543.
Solution 2
To find the p-value, we first need to calculate the z-score. The z-score is calculated using the formula:
z = (X - μ) / (σ / √n)
where: X = sample mean = 540 μ = population mean = 543 σ = standard deviation = 10 n = sample size = 30
Substituting the values into the formula, we get:
z = (540 - 543) / (10 / √30) = -1.732
Since we are testing whether the population mean is less than 543, we are dealing with a one-tailed test. The p-value is the probability that we would observe a z-score as extreme as -1.732 or more so, assuming the null hypothesis is true.
Looking up the z-score of -1.732 in a standard normal distribution table or using a calculator, we find that the corresponding probability is 0.0418.
However, since we are conducting a one-tailed test at a significance level of 0.01, we need to halve this probability to find the p-value. Therefore, the p-value is 0.0418 / 2 = 0.0209.
So, the p-value is 0.0209.
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