Asample of 56 research cotton samples resulted in a sample average percentage elongation of 8.63 and a sample standard deviation of 0.79. calculate a 95% large-sample ci for the true average percentage elongation μ. (give answers accurate to 2 decimal places.)

To solve for the confidence interval for the true average percentage elongation, we use the z statistic. The formula for confidence interval is given as:

Confidence interval = x ± z σ / sqrt (n)

where,

x = the sample mean = 8.63

σ = sample standard deviation = 0.79

n = number of samples = 56

From the standard distribution tables, the value of z at 95% confidence interval is:

z = 1.96

Therefore substituting the known values into the equation:

Confidence interval = 8.63 ± (1.96) (0.79) / sqrt (56)

Confidence interval = 8.63 ± 0.207

Confidence interval = 8.42, 8.84

correct answer is c. just did it

step-by-step explanation: