Question

In a sample of 12 participants, a researcher estimates the 80% CI for a sample with a mean of M = 22.3 and an estimated standard error (SM) of 4.7. What is the confidence interval at this level of confidence? 80% CI 12.1, 32.5 80% CI 17.6, 27.0 80% CI 15.9, 28.7 There is not enough information to answer this question.

Answer 1

Answer: 80% CI (15.9, 28.7).

Step-by-step explanation:

We have given that ,

Sample size : n = 12

Confidence interval : 80% = 0.80

Significance level : 1-0.80=0.20

Sample mean : $\overline{x}$

Estimated standard error : SE = 4.7

Degree of freedom : df = n-1 =11

By t-distribution table , the critical t-value for df = 11 and significance lvel of 0.20 = $t*=1.3634$

Then, the 80% confidence interval will be :

$\overline{x}\pm t^*SE\\\\=22.3\pm (1.3634)(4.7)\\\\=22.3\pm6.40798\\\\ (22.3-6.40798,\ 22.3+6.40798)\\\\=(15.89202,\ 28.70798)\approx(15.9\ ,\ 28.7)$

Hence, the correct option is 80% CI (15.9, 28.7).