To find the 90% confidence interval for the mean, we will use the formula:
CI = X̄ ± Z * (σ/√n),
where X̄ is the sample mean, Z is the z-score corresponding to the desired confidence level (90% in this case), σ is the population standard deviation (assuming it is known), and n is the sample size.
First, we calculate the sample mean (X̄) and the sample standard deviation (s):
X̄ = (54 + 56 + 59 + 59 + 65 + 66 + 67 + 68 + 69 + 76 + 84 + 86 + 89 + 94 + 101) / 15
= 994 / 15
≈ 66.267
Next, we calculate the sample standard deviation (s):
s = √((∑(Xi - X̄)²) / (n-1))
= √((∑(54-66.267)² + (56-66.267)² + (59-66.267)² + ... + (101-66.267)²)) / (15-1))
≈ √(13200.533) / 14
≈ √942.895
≈ 30.710
Assuming the population standard deviation is known, we can find the z-score corresponding to a 90% confidence level (α = 0.10 or 1-α = 0.90). Using a standard normal distribution table, we find the corresponding z-score to be approximately 1.645.
Finally, we can calculate the confidence interval:
CI = 66.267 ± 1.645 * (30.710/√15)
≈ 66.267 ± 1.645 * (30.710/√15)
≈ 66.267 ± 1.645 * (7.931)
≈ 66.267 ± 13.048
Therefore, the 90% confidence interval for the mean is approximately (53.219, 79.315) rounded to three decimals.
Assuming CLT applies for the sample data given by 54, 56, 59, 59, 65, 66, 67, 68, 69, 76, 84, 86, 89, 94, 101 Find a 90% CI to three decimals for the mean
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