Question

df to.10 0.025 t0.01 to.0ns df Values of ta 3.078 6.314 2.706 31.821 63.657 1886 2.920 4.303 6.965 9.925 1.638 2.353 3.182 .54 5.841 1.533 2.132 2.776 3.747 4.604 0 1.476 2.015 2.5 3.365 4.032 1.440 1.93 2.4473.143 3.707 1.415 1895 2.365 2.998 3.499 1.397 1860 2.306 2.896 3.355 1.383 1833 2.22 2.82 3.250 1.372 1.812 .282.764 3.169 1.363 1.796 2.20 2.718 3.106 1.356 1.782 2.179 2.68 3.055 1.350 77 2.160 2.650 3.012 1.345 76 2.145 2.624 2.977 10 10 12 13 14 15 16 17 18 19 20 21 12 13 14 15 16 17 18 19 20 21 1.341 1.753 2.3 2.602 2.947 1.337 1746 2.120 2.583 2.921 1.333 1.740 2.110 2.567 2.898 1.330 734 2.10 2.552 2.878 1.328 1.729 2.093 2.539 2.861 1.325 1725 2.086 2.528 2.845 1.323 172 2.080 2.518 2.83 1.32 1717 2.074 2.508 2.819 1.319 71 2.069 2.500 2.807 1.318 7 2.4 2.492 2.797 23 24 23 24

Answer 1

Given :-

Sample mean ( $\large \bar X$ ) = 2.0

Sample standard deviation (s) = 4.7

Sample size (n) = 61

Degree of freedom (df) =n-1 =61-1 =60

Confidence level (c) = 98%

Then critical value is t_​​​c= 2.39

Now, we have to find confidence interval for mean

Therefore,

  $\large \mu$ = ( $\large \bar X \pm \ t_c \ * (s/\sqrt n )$ )

$\large \mu$ = (  $\large \2.0 \pm \ 2.39 \ * (4.7/\sqrt 61 )$ )

$\large \mu$ = ( 2.0 - 1.44 , 2.0 + 1.44 )

  $\large \mu$​​​​​​ = ( 0.56 , 3.44 )

Therefore the 98% confidence interval about mean ($\large \mu$) is 0.56 to 3.44 .