Question

Recorded here are the germination times (in days) for twenty randomly chosen seeds of a new type of bean: 18, 12, 11, 21, 16, 20, 12, 19, 21, 16, 21, 15, 13, 17, 20, 15, 14, 10, 14, 16 Send data to Excel Assuming that germination times are normally distributed, find a 95% confidence interval for the mean germination time for all beans of this type. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas) What is the lower limit of the confidence interval? What is the upper limit of the confidence interval?

Answer 1

The mean of the sample is calculated as

$M=\frac{X1+X2+....X20}{20}=16.5$

And sample standard deviation as

S= 4.478

Now.,

The formula for confidence interval of a normal distribution estimation is:

μ = M ± t(sM)

where:

M = sample mean
t = t statistic determined by confidence level
sM = standard error = √(s2/n)

Calculation

M = 16.5
t = 2.09, t is calculated using t distribution table or calculator at 95% confidence level
sM = √(4.4782/20) = 1

μ = M ± t(sM)
μ = 16.5 ± 2.09*1
μ = 16.5 ± 2.096

95% CI [14.404, 18.596].

Lower limit= 14.404

Upper limit= 18.596