The null and alternate hypotheses are:
H0: μ1 = μ2
H1: μ1 ≠ μ2
A random sample of 15 items from the first population showed a mean of 50 and a standard deviation of 5. A sample of 12 items for the second population showed a mean of 46 and a standard deviation of 15. Assume the sample populations do not have equal standard deviations.
Do you understand the difference between a sample and a population?
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score and compare it to whatever level of significance you are using.
To test the null and alternative hypotheses, we can use the two-sample t-test since we have samples from two different populations.
The formula for the two-sample t-test is:
t = (x1 - x2) / sqrt((s1^2 / n1) + (s2^2 / n2))
where:
- t is the test statistic
- x1 and x2 are the sample means
- s1 and s2 are the sample standard deviations
- n1 and n2 are the sample sizes
Now let's calculate the values using the provided data:
x1 = 50
s1 = 5
n1 = 15
x2 = 46
s2 = 15
n2 = 12
t = (50 - 46) / sqrt((5^2 / 15) + (15^2 / 12))
First, let's calculate the denominators:
denom1 = s1^2 / n1 = 5^2 / 15
denom2 = s2^2 / n2 = 15^2 / 12
denom1 = 25 / 15 = 1.67
denom2 = 225 / 12 = 18.75
Now substitute the values into the t formula:
t = (50 - 46) / sqrt(1.67 + 18.75)
t = 4 / sqrt(20.42)
t = 4 / 4.52
t ≈ 0.885
Now, we need to determine the critical value for the desired level of significance (alpha). Let's assume alpha is 0.05, for a 95% confidence level. Since it is a two-tailed test, we need to divide alpha by 2, giving us 0.025.
Using a t-table or a statistical software, we find that for a degrees of freedom (df) of 12 + 15 - 2 = 25 and alpha = 0.025, the critical t-value is approximately ± 2.064.
Finally, we compare the calculated t-value with the critical t-value to make a decision:
If |t| > critical t-value, then reject the null hypothesis (H0).
If |t| ≤ critical t-value, then fail to reject the null hypothesis (H0).
Since |0.885| ≤ 2.064, we fail to reject the null hypothesis.
Therefore, based on the given data, there is not enough evidence to suggest a significant difference between the means of the two populations.