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).
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.
2 answers
Correction:
Your calculation is correct.
z = 5.156
You would still reject the null and accept the alternate hypothesis.
Sorry for any confusion.
Your calculation is correct.
z = 5.156
You would still reject the null and accept the alternate hypothesis.
Sorry for any confusion.
1 answer