Answer:
97% Confidence interval = (12.62, 18.98)
Step-by-step explanation:
Complete Question
Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean: 18, 12, 20, 17, 14, 15, 13, 11, 21, 17. Assume that the population germination time is normally distributed. Find the 97% confidence interval for the mean germination time.
Solution
We first compute the sample mean and standard deviation for this sample distribution
Sample mean = (Σx)/N = (158/10) = 15.8
Standard deviation = √{[Σ(x - xbar)²]/(N-1)} = 3.3598941782278 = 3.36
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 = 15.8
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 = 10 - 1 = 9
Significance level for 97% confidence interval
(100% - 97%)/2 = 1.5% = 0.015
t (0.015, 9) = 2.9982 (from the t-tables)
Standard error of the mean = σₓ = (σ/√n)
σ = standard deviation of the sample = 3.36
n = sample size = 10
σₓ = (3.36/√10) = 1.0625
97% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]
CI = 15.8 ± (2.9982 × 1.0625)
CI = 15.8 ± 3.1809882246
97% CI = (12.6190117754, 18.9809882246)
97% Confidence interval = (12.62, 18.98)
Hope this Helps!!!
