In 2002, the average price of new homes in a certain suburb was $145,000. Assume that this mean is based on a random sample of 1000 new home sales and that the sample standard deviation is $24,000. Construct a 95% confidence interval for the 2002 mean price of all such homes
95% = mean ± 1.96 SEm
SEm = SD/√n
To construct a 95% confidence interval for the 2002 mean price of all new homes in the suburb, we will be using the formula:
Confidence Interval = sample mean ± (critical value) * (sample standard deviation / sqrt(sample size))
Step 1: Identify the known values from the problem:
- Sample mean: $145,000
- Sample standard deviation: $24,000
- Sample size: 1000
- Confidence level: 95%
Step 2: Find the critical value corresponding to a 95% confidence level. This critical value is based on the z-score for the standard normal distribution. The most common value for a 95% confidence level is 1.96. This can be obtained from a z-table or calculated using statistical software.
Step 3: Calculate the standard error, which is the sample standard deviation divided by the square root of the sample size:
Standard Error = sample standard deviation / sqrt(sample size)
Step 4: Substitute the known values into the formula:
Confidence Interval = $145,000 ± (1.96) * ($24,000 / sqrt(1000))
Step 5: Simplify the equation:
Confidence Interval = $145,000 ± (1.96) * ($24,000 / 31.62)
Step 6: Calculate the values:
Confidence Interval = $145,000 ± (1.96) * $760.25
Step 7: Calculate the upper and lower bounds of the confidence interval:
Upper bound = $145,000 + (1.96) * $760.25
Lower bound = $145,000 - (1.96) * $760.25
Step 8: Calculate the actual values:
Upper bound = $147,509.79 (rounded to nearest cent)
Lower bound = $142,490.21 (rounded to nearest cent)
Therefore, we can say with 95% confidence that the mean price of all new homes in the suburb in 2002 falls between $142,490.21 and $147,509.79.