Question

1. Suppose we wish to estimate the mean heart rate, using a 95% confidence interval, for a par- ticular population. We observe 130 individuals to see a sample mean of 98.249 and a sample standard deviation of 0.733. We then find the tc-value, corresponding to the value on the t-distribution (d.f. = 130) so that the area to the left of it is equal to 95%, which ends up being:

   tc ≈ 1.656659.
   Next we compute the error and obtain our confidence interval of

   98.1425 < μ < 98.355. What horrible, horrible, mistake(s) have we made?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 98.249

sample standard deviation = s = 0.733

sample size = n = 130

Degrees of freedom = df = n - 1 = 130 - 1 = 129

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,129 = 1.979

Margin of error = E = t/2,df * (s /n)

= 1.979 * ( 0.733 / 130)

Margin of error = E = 0.127

The 95% confidence interval estimate of the population mean is,

- E < < + E

98.249 - 0.127 < < 98.249 + 0.127

( 98.122 < < 98.376 )