suppose a large Labor union wishes to estimate the mean number of hours per month a union member is absent from work. The union decides to sample 120 of its members at random and monitor they’re working time for one month. At the end of the month, the total number of hours absent from the work is required for each employee. If the mean and standard deviation of the sample or 9.6 hours and 6.4 hours respectively find a 95% confidence interval for for the true mean number of hours absent per month per employee. round to four decimal places
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Step 1
Given: Sample mean (x̄) = 9.6 Standard deviation (σ) = 6.4 Sample size (n) = 120 Confidence level = 95% Critical value (z) for 95% confidence level = 1.96 Margin of error (E) = z * (σ / √n) E = 1.96 * (6.4 / √120) E = 1.96 * (6.4 / 10.95445) E = 1.96 * 0.5847 E = Show more…
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