Answer:
99% Confidence interval for the population mean noise level = (110.8, 151.2) in decibels
Explanation:
Noise levels = 144,119,140,143,128,112
Sample mean = xbar = (sum of all data)/(sample size)
Sample mean = (Σx)/N = (144 + 119 + 140 + 143 + 128 + 112)/6 = 131.0 to 1 d.p
Standard deviation = √[Σ(x - xbar)²/N]
Σ(x - xbar)² = [144 - 131]² + [119 - 131]² + [140 - 131]² + [143 - 131]² + [128 - 131]² + [112 - 131]²
= 169 + 144 + 81 + 144 + 9 + 361 = 908
Standard deviation = √(908/6) = 12.30.
Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.
Mathematically,
Confidence Interval = (Sample mean) ± (Margin of error)
Sample Mean = 131 dB
Margin of Error is the width of the confidence interval about the mean.
It is given mathematically as,
Margin of Error = (Critical value) × (standard Error of the mean)
Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.
To find the critical value from the t-tables, we first find the degree of freedom and the significance level.
Degree of freedom = df = n - 1 = 6 - 1 = 5.
Significance level for 99% confidence interval
(100% - 99%)/2 = 0.5% = 0.005
t (0.005, 5) = 4.032 (from the t-tables)
Standard error of the mean = σₓ = (σ/√n)
σ = standard deviation of the sample = 12.30
n = sample size = 6
σₓ = (12.30/√6) = 5.02
99% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]
CI = 131 ± (4.032 × 5.02)
CI = 131 ± 20.24064
99% CI = (110.75936, 151.24064)
99% Confidence interval = (110.759, 151.241)
Hope this Helps!!!