Hi,
You would need to perform a one-sample test to establish if the sample mean of $674,000 is significantly different from the population mean of $650,000 at the alpha level of 2%. You can either use the one-sample z test or the one-sample t test. I present summaries of both below (both using the p-value and critical-value approaches).
One-sample z test
Z Formula: (sample mean - population mean)/[standard deviation/sqrt(n)]
Current example: (674,000 - 650,000)/[94,500/sqrt(70)]
In this example, Z would equal 2.1249. To determine the p-value of this Z score, you could use a Z table or calculator online. I found a p-value of .033595. Because this p-value is not less than .02, we would conclude that the sample mean is not significantly different from the population mean. It was not stated in your question, so I assumed that a two-tailed hypothesis test was used. Important note: if a one-tailed test was used, the p-value would be divided by two (.033595/2 = 0.0167975) and would reach significance.
To use the critical-value method, you would have to estimate the 98% confidence interval around the sample mean and see if it overlaps with the population mean or not. It is the 98% confidence interval because alpha was set at the 2% level (100% - 2% = 98%).
Confidence Interval formula: sample mean +/- Zcrit * [standard deviation/sqrt(n)]
Current example: 674,000 +/- 2.326 * [94,500/sqrt(70)]
In this example, the 98% confidence interval would range from $647,728 to $700,272. Because this confidence interval includes the population mean (i.e., $650,000), you would conclude that the sample mean is not significantly different from the population mean.
One-sample t test
The t test methods are similar to the one-sample z test above with the following changes. The t formula is the same as the Z formula, but you use a t chart to determine if the value is significant or not. You need to know what the degrees of freedom of the sample is. The df is simply the sample size minus 1 here, so the df would be 69. I got a p-value of .037187, so because it is not less than .02 you would conclude the difference is not significant.
The critical-value method is also similar except you have to substitute Tcrit in for Zcrit.:
Confidence Interval formula: sample mean +/- Tcrit * [standard deviation/sqrt(n)]
Current example: 674,000 +/- 2.382 * [94,500/sqrt(70)]
In this example, the 98% confidence interval would range from $647,095.52 to $700,904.48. Because this confidence interval includes the population mean (i.e., $650,000), you would conclude that the sample mean is not significantly different from the population mean.
Hope this explanation and example helps! Let me know if you have any other questions.
-Mike
