Question

Identify distribution and find a (1-α) 100% confidence interval for a population mean μ for these values:

α = .10, n = 65, x̄=1049, s^2 = 51

We are ___ % confident that the mean is between ___ and ___.

Please show your work, thank.

Answer 1

Answer)

As here population standard deviation is unknown, and we are given with sample variance

Therefore we will use t distribution

In case we are given with population standard deviation that is sigma, we use z table

But here we will use t distribution

Given alpha = 0.10

Confidence level = (1-alpha)*100

= (1-0.1)*100 = 90%

From t table, for degrees of freedom n-1, 65-1 = 64 and 90% confidence level

t = 1.669013

margin of error = t*(s.d/√n)

= 1.669013*(√51/√65)

As standard deviation = square root of variance

= 1.47838698016

Confidence interval is given by

Lower bound = mean - moe = 1049-1.47838 = 1047.52161301

Upper bound = mean + moe = 1049-1.47838 = 1050.47838698

(1047.52, 1050.48)

We are 90% confident that true population mean lies in between 1047.52 and 1050.48