According to the local real estate board, the average number of days that homes stay on the market before selling is 78.4 with a standard deviation equal to 11 days. A prospective seller selected a random sample of 36 homes from the multiple listing service. Above what value for the sample mean should 95 percent of all possible sample means fall?
78.4-+ 1.96*11/sqrt(36))
(74.81, 81.99)
78.4-+ 1.96*11/sqrt(36))
(74.81, 81.99)
To determine the value above which 95% of all possible sample means should fall, we need to calculate the margin of error. The margin of error can be determined using the formula:
Margin of Error = (Z * standard deviation) / √(sample size)
Where Z represents the Z-score corresponding to the desired level of confidence. For a 95% confidence level, the Z-score is approximately 1.96.
In this case, the standard deviation is given as 11 days, and the sample size is 36.
Margin of Error = (1.96 * 11) / √(36)
Calculating this, we get:
Margin of Error ≈ 3.618
To find the value above which 95% of all possible sample means should fall, we add the margin of error to the average number of days homes stay on the market:
Sample mean + Margin of Error = 78.4 + 3.618 = 82.018
Therefore, above the value of 82.018 for the sample mean, 95% of all possible sample means would fall.
To find the value above which 95 percent of all possible sample means should fall, we need to use the concept of the sampling distribution of the sample mean.
The sampling distribution of the sample mean is approximately normally distributed, with a mean equal to the population mean (78.4 days) and a standard deviation equal to the population standard deviation divided by the square root of the sample size.
In this case, the population standard deviation is 11 days, and the sample size is 36 homes. So, the standard deviation of the sampling distribution is 11 / sqrt(36) = 11 / 6 = 1.833 days.
To find the value above which 95 percent of all possible sample means should fall, we can utilize the concept of the z-score. The z-score represents the number of standard deviations a particular value is away from the mean.
Since we want to find the value above which 95 percent of all possible sample means fall, we need to find the z-score that corresponds to the upper 5 percent of the distribution. In a standard normal distribution (a normal distribution with mean 0 and standard deviation 1), this z-score is equal to 1.645.
To find the corresponding value in the actual distribution, we can multiply the z-score by the standard deviation of the sampling distribution and add it to the mean of the distribution:
Value = Mean + (Z-score * Standard Deviation)
= 78.4 + (1.645 * 1.833)
= 78.4 + 3.018285
= 81.418285
Therefore, above 81.418285, 95 percent of all possible sample means should fall.