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To estimate the average monthly income of workers in ascertained factory a sample of 100 workers was taken with a mean of 400 birr and a standard deviation of 20. Find a 90 percent confidence interval for the population means and interpret the result

User Toshihiko
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Answer:

To find a 90 percent confidence interval for the population mean of the workers' monthly income in the factory, you can use the formula:

Confidence interval = sample mean ± (critical value * standard error)

First, you need to find the critical value corresponding to a 90 percent confidence level. Since the sample size is large (n = 100), we can use the Z-distribution. For a 90 percent confidence level, the critical value is approximately 1.645.

Next, you calculate the standard error, which is the standard deviation of the sample divided by the square root of the sample size:

Standard error = standard deviation / √(sample size)

Standard error = 20 / √(100)

Standard error = 20 / 10

Standard error = 2

Now you can substitute the values into the confidence interval formula:

Confidence interval = 400 ± (1.645 * 2)

Calculating the confidence interval:

Lower bound = 400 - (1.645 * 2) = 400 - 3.29 = 396.71

Upper bound = 400 + (1.645 * 2) = 400 + 3.29 = 403.29

The 90 percent confidence interval for the population mean of the workers' monthly income is approximately 396.71 to 403.29 birr.

Step-by-step explanation:

Interpretation: We are 90 percent confident that the true population mean of the workers' monthly income in the factory falls within the range of 396.71 to 403.29 birr. This means that if you were to take multiple samples and calculate confidence intervals for each sample, approximately 90 percent of those intervals would contain the true population mean.

User Meir
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