Answer:
a) variable = X
N = 15
Mean = 50.0933
SE Mean = 1.58
StDev = 6.11
Variance = 37.3321
Sum = 751.40
b) 95% Confidence interval = (46.997, 53.190) = (47.00, 53.19)
Explanation:
variable = X
N = ?
Mean = ?
SE Mean = 1.58
StDev = 6.11
Variance = ?
Sum = 751.40
To fill in the missing quantities, we need to use some of their formulas.
For N, the number of variables
SE = Standard Error of the mean = σₓ = (σ/√N)
σₓ = Standard Error of the mean = 1.58
σ = standard deviation of the sample = 6.11
N = sample size = ?
1.58 = (6.11/√N)
√N = (6.11/1.58) = 3.8671
N = 3.8671² = 14.954374299 = 15 to the nearest whole number.
For the Mean
Mean = (Sum of variables)/(Number of variables)
Mean = ?
Sum of variables = 751.40
Number of variables = N = 15
Mean = (751.40/15) = 50.0933333333 = 50.093
For Variance
Variance = (Standard deviation)² = (6.11)² = 37.3321
b) To compute the confidence interval.
idence 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 = 50.0933
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 z-distribution. This is because although, there is no information provided for the population standard deviation, the sample size is large enough for the sample properties to approximate the population properties.
To find the critical value from the z-tables,
z at 95% confidence level = 1.960 (from the z-distribution tables)
Standard error of the mean = σₓ = (σ/√N)
Already calculated to be 1.58
95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]
CI = 50.0933 ± (1.960 × 1.58)
CI = 50.0933 ± 3.0968
95% CI = (46.9965, 53.1901)
95% Confidence interval = (46.997, 53.190) = (47.00, 53.19)
Hope this Helps!!!