Final answer:
The 95% confidence interval for the difference between the mean home prices in two areas, with known population standard deviations, is approximately $6,876 to $37,524, rounded to the nearest whole number.
Step-by-step explanation:
To construct a 95% confidence interval for the true difference between the mean home prices in two areas, we need to use the formula for the confidence interval of the means for two independent populations when the population standard deviations are known. The formula for the confidence interval is:
CI = (x1 - x2) ± Z*(√(σ²_1/n1 + σ²_2/n2))
where x1 and x2 are the sample means of the two populations, σ1 and σ2 are the population standard deviations, n1 and n2 are the sample sizes, and Z* is the Z-value that corresponds to the desired level of confidence. For a 95% confidence interval, the Z-value is 1.96.
Using the given information:
- Mean home price in the first area (x1) = $190,500
- Standard deviation in the first area (σ1) = $33,975
- Number of homes sampled in the first area (n1) = 40
- Mean home price in the second area (x2) = $168,300
- Standard deviation in the second area (σ2) = $32,670
- Number of homes sampled in the second area (n2) = 33
The confidence interval calculation is as follows:
CI = (190,500 - 168,300) ± 1.96*(√((33,975)²/40 + (32,670)²/33))
After calculating the values inside and outside of the square root, we find the margin of error and therefore the confidence interval:
CI = 22,200 ± 1.96*(√((1,150,502,500)/40 + (1,067,684,900)/33))
CI = 22,200 ± 1.96*(√(28,762,562.5 + 32,354,390.91))
CI = 22,200 ± 1.96*(√(61,116,953.41))
CI = 22,200 ± 1.96*(7,818.16)
CI = 22,200 ± 15,324.39
CI = (6,875.61, 37,524.39)
Therefore, the 95% confidence interval for the difference between the mean home prices is approximately $6,876 to $37,524, rounded to the nearest whole number.