Suppose the mean income of 35-year-olds in the US is $25,000. A random sample of 150 35-year-olds in California results in a sample mean income of $26,600 and a sample standard deviation of $3800. At the 5% significance level, ca we conclude that the true mean income of 35-year-old Californians is greater than that of the nation, in general?
the answer I have gotten was 5.156, right tailed.
Formula:
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
With your data:
z = (26600 - 25000)/(3800/√150) = 4.96
You can reject the null and accept the alternate hypothesis (µ > 25000).
Correction:
Your calculation is correct.
z = 5.156
You would still reject the null and accept the alternate hypothesis.
Sorry for any confusion.
To determine whether we can conclude that the true mean income of 35-year-old Californians is greater than that of the nation, we need to conduct a hypothesis test.
Let's set up our null and alternative hypotheses:
- Null hypothesis (H0): The true mean income of 35-year-old Californians is not greater than the national mean income (μ ≤ $25,000).
- Alternative hypothesis (Ha): The true mean income of 35-year-old Californians is greater than the national mean income (μ > $25,000).
To calculate the test statistic, we will use the formula for a t-test:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
Given the information, we have:
- Sample mean (x̄) = $26,600
- Population mean (μ) = $25,000
- Sample standard deviation (s) = $3,800
- Sample size (n) = 150
Substituting these values into the formula, we get:
t = ($26,600 - $25,000) / ($3,800 / sqrt(150))
Calculating this expression, we find t ≈ 5.156.
Since we're performing a right-tailed test with a 5% significance level, we compare the calculated t-value to the critical t-value. We need to find the critical t-value with (n-1) degrees of freedom (df = 150-1 = 149) and a significance level of 5% for a one-tailed test.
Using a t-table or statistical software, we find that the critical t-value at a 5% significance level and 149 degrees of freedom is approximately 1.658.
Since the calculated t-value (5.156) is greater than the critical t-value (1.658), we reject the null hypothesis.
Therefore, we can conclude that the true mean income of 35-year-old Californians is greater than the national mean income at a 5% significance level.