The question is asking to calculate the probability that the average income of 36 independent real estate agents in Canada has been greater than $130,000 given that the population of real estate agent income is normally distributed with a mean of $144,300 and a “standard deviation” of $32,450.
|The central limit theorem states that the sample mean of a given set of data has a normal distribution with mean μ and “standard deviation” σ/√n, where n is the sample size. This theorem can be used to approximate the distribution of the sample mean of a given data.
In this case, the mean is mu = $144,300, and the “standard deviation” is σ = $32,450. Since a sample size of n = 36, it can be calculated that “the standard error of the mean” as:
“Standard error” = σ/√n = $32,450/√36 = $5,408.33
The z-score or the number of “standard deviations” from the sample mean that deviates from the population mean, must be calculated in order to ascertain the chance that the sample mean has been more than $130,000. The likelihood of receiving a big z-score may then be calculated or found using a conventional normal distribution table (Al-Amshawee et al., 2020).
z-score = (sample mean – population mean) / standard error = ($130,000 – $144,300) / $5,408.33 = -2.642
Using the “standard normal distribution table”, it can be found that the “probability of a z-score” being less than -2.642 is 0.0042. Therefore, “the probability” that the mean of the sample exceeds $130,000 is approximately:
P (sample mean > $130,000) = 1 – P (sample mean ≤ $130,000)
= 1 – P (z-score ≤ -2.642)
= 1 – 0.0042
Therefore, the probability that the mean of the sample exceed $130,000 is approximately 0.9958 or 99.58%.
The central limit theorem is a statistical concept that helps to approximate the distribution of the average of a sample. This average, which is equal to the population mean, is calculated by dividing the population “standard deviation” by the square of the sample size. The central limit theorem states that the sampling distribution of the mean is approximately normally distributed. The formula z = (x -) / SE is then used to calculate the sample mean’s z-score, where x denotes the sample mean, p denotes “the population means”, and se is the “standard error” of the mean (Ali et al., 2020).
The sample mean was found to have a z-score of -2.647, which means that it was 2.647 standard errors below “the population means”. The probability of a z-score less than -2.647 was also found to be around 0.004, or 0.4%, using a standard normal distribution table or calculator.
As a result, there is a 99.6% chance that “the sample’s mean” income has been higher than $130,000 (1 – 0.004 = 0.996). Given the sample size and population distribution characteristics, there is a high likelihood that the sample mean exceeds $130,000.
|In order to calculate a confidence interval for the average amount of gallons of petrol purchased each visit to a petrol station, it is needed to know the sample mean, “sample standard deviation”, sample size, and degree of confidence.
Sample mean (x̄) = (8.7+22.4+9.5+13.3+18.9+22+14.4+35.7+19+24.9+5.7+15.7+8.9+22.5+15.9)/15 = 17.02
“Sample standard deviation”, (s) = “sqrt ([∑ (xᵢ – x̄) ²]/(n-1)) = sqrt ([(8.7-17.02) ² + (22.4-17.02) ² + … + (15.9-17.02) ²]/ (15-1)) = 7.035”
Sample size (n) = 15
Level of confidence = 95% = 0.95
It is suggested to use the t-distribution rather than the standard normal distribution when dealing with a small sample size (n 30). The t-distribution has 14 degrees of freedom (n-1) and the critical value of 2.145 is necessary for a two-tailed test with a 95% confidence level (Bruns, 2019).
The margin of error can be calculated as follows:
Margin of error = t_critical * (s/sqrt(n)) = 2.145 * (7.035/sqrt (15)) = 4.23
Therefore, “the 95% confidence interval for the population means gallons of gasoline” purchased per visit to a gasoline station is:
17.02 ± 4.23 Or (12.79, 21.25)
It can be 95% confident that the actual population means gallons of gasoline purchased per visit to a gasoline station falls between 12.79 and 21.25 (Faraj et al. 2020).
A confidence interval is an estimated value derived from sample data that gives a value for uncertainty or degree of confidence in the estimate’s correctness. Based on the sample data and the degree of confidence chosen, the confidence interval represents the feasible value range within which the population means is likely to fall. In this instance, the calculations are focused on calculating the typical amount of petrol Canadians buy each time they visit a petrol station. This is accomplished by taking a random sample of 15 clients and their associated petrol purchase amounts. To create a confidence interval, the model mean and “standard error” must be known. The sample mean is calculated as the total of all sample values divided by the sample size. The standard error, on the other hand, measures how variable the mean estimate is (Druckman, & McGrath, 2019). “The sample standard deviation” is used as a rough approximation when the “population standard deviation” is unknown.
Figure 1: Summary Output
(Source: Acquired from Excel)
Figure 2: Coefficient and standard error
(Source: Acquired from Excel)
Figure 3: Analysis of Z-score table
(Source: Acquired from Excel)
The sample mean is determined to be 16.18 gallons using “the sample data supplied”, and “the sample standard deviation” is determined to be 7.49 gallons. It is necessary to determine the crucial value from the t-distribution table in order to build a 95% confidence interval. As there are 15 samples, 14 “degrees of freedom” are assumed. “The critical value with a 95% confidence level” and 14 “degrees of freedom” is 2.145, according to the t-distribution table (Chen et al., 2020).
The following formula is used to determine the confidence interval:
Critical value * = confidence interval = sample mean (standard error)
The following formula is used to determine “the standard error”:
“The standard error” is calculated as “sample standard deviation” / sqrt (sample size)
The result of plugging in the values is:
“Standard deviation”: sqrt (15) = 1.93 / 7.49
Confidence interval = 16.18 ± (2.145) * (1.93) = (12.38, 19.98)
Thus, the computation gives “95% confidence” that “the actual population mean range” for the amount of petrol that Canadians purchase every visit to a petrol station is between 12.38 and 19.98 gallons.
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