A random sample of 50 households in community A has a mean household income of X- =Rs. 44,600 with a standard deviation s =Rs. 2,200. A random sample of 50 households in community B has a mean of _X = Rs. 43,800 with a standard deviation of Rs. 2,800. Estimate the difference in the average houshold income in the two communities using a 95 percent confidence interval?
To estimate the difference in the average household income between two communities, we can use the two-sample t-test and construct a confidence interval. Here's how you can do it:
Step 1: Identify the necessary information:
Sample size (n1) for community A = 50
Sample mean (X1) for community A = Rs. 44,600
Standard deviation (s1) for community A = Rs. 2,200
Sample size (n2) for community B = 50
Sample mean (X2) for community B = Rs. 43,800
Standard deviation (s2) for community B = Rs. 2,800
Confidence level = 95% (which corresponds to alpha value of 0.05)
Step 2: Calculate the standard error of the difference:
Standard error (SE) of the difference in means = sqrt[(s1^2/n1) + (s2^2/n2)]
SE = sqrt[(2200^2/50) + (2800^2/50)]
Step 3: Compute the t-value:
Degrees of freedom (df) = (n1 - 1) + (n2 - 1) = 48 + 48 = 96
t-value = critical value at alpha/2 for the given degrees of freedom (df)
Step 4: Calculate the margin of error (MoE):
MoE = t-value * SE
Step 5: Calculate the lower and upper bounds for the confidence interval:
Lower bound = (X1 - X2) - MoE
Upper bound = (X1 - X2) + MoE
Let's calculate the confidence interval:
Standard error (SE) = sqrt[(2200^2/50) + (2800^2/50)] ≈ Rs. 392.61
Now, we need to find the t-value for a 95% confidence level:
t-value ≈ 1.984 (using a t-table or statistical software)
Margin of error (MoE) = t-value * SE ≈ 1.984 * 392.61 ≈ Rs. 777.58
Lower bound = (X1 - X2) - MoE = (44600 - 43800) - 777.58 ≈ Rs. 22.42
Upper bound = (X1 - X2) + MoE = (44600 - 43800) + 777.58 ≈ Rs. 1522.42
Therefore, the 95% confidence interval estimate for the difference in average household income between the two communities is approximately Rs. 22.42 to Rs. 1522.42.
To estimate the difference in the average household income in the two communities (community A and community B) using a 95 percent confidence interval, we can use the formula for the confidence interval for the difference between two means.
The formula for the confidence interval for the difference between two means is:
CI = (_X1 - _X2) ± t * SE
where:
- CI refers to the confidence interval
- _X1 and _X2 are the sample means for the two communities
- t is the critical value from the t-distribution based on the desired confidence level and the degrees of freedom
- SE is the standard error of the difference in means, given by the formula:
SE = sqrt((s1^2 / n1) + (s2^2 / n2))
where:
- s1 and s2 are the sample standard deviations for the two communities
- n1 and n2 are the sample sizes for the two communities
Given the following information:
For community A:
- _X1 = Rs. 44,600
- s1 = Rs. 2,200
- n1 = 50
For community B:
- _X2 = Rs. 43,800
- s2 = Rs. 2,800
- n2 = 50
Now, we can calculate the confidence interval.
Step 1: Calculate the standard error (SE)
SE = sqrt((2200^2 / 50) + (2800^2 / 50))
= sqrt((4840000/50) + (7840000/50))
= sqrt(96800 + 156800)
= sqrt(253600)
= 503.58
Step 2: Calculate the critical value (t)
Since the sample size for both communities is 50, the degrees of freedom will be (50 + 50 - 2) = 98.
At a 95 percent confidence level, the critical value from the t-distribution for a two-tailed test with 98 degrees of freedom is approximately 2.626.
Step 3: Calculate the confidence interval (CI)
CI = (44600 - 43800) ± 2.626 * 503.58
= 800 ± 1319.97
= (800 - 1319.97, 800 + 1319.97)
≈ (-519.97, 2119.97)
Therefore, the 95 percent confidence interval for the difference in the average household income between the two communities is approximately (-Rs. 519.97, Rs. 2119.97).