Question

The principal randomly selected six students to take an aptitude test. Their scores were: 81.6 72.0 81.1 86.4 70.2 83.1 Determine a 90% confidence interval for the mean score for all students

Answer 1

Answer:

84.38, 73.74

Step-by-step explanation:

score given 81.6, 72.0, 81.1, 86.4, 70.2, 83.1

sample size (n) = 6

$mean = \dfrac{81.6+ 72.0+ 81.1+ 86.4+ 70.2+ 83.1}{6}$

mean = 79.06

standard deviation

$\sigma =\sqrt{\frac{\sum (x-\bar{x})^2}{n-1}}$

$\sigma =\sqrt{\frac{ (81.6-79.06)^2+(72-79.06)^2+(86.4-79.06)^2+(70.2-79.06)^2+(81.1-79.06)^2+(83.1-79.06)^2}{6-1}}$

σ = 6.47

level of significance (α) = 1 - 90% = 10%

confidence interval

$\bar{x} \pm t_{\alpha}(\frac{S}{\sqrt{n}})\\79.06 \pm 2.015(\frac{6.47}{\sqrt{6} })$

=79.06 ± 5.32

= 84.38, 73.74