Answer:
a) (17.227, 22.773)
b) 2.773
c) 2.773
Step-by-step explanation:
Given:
Sample size, n = 25
Standard deviation, s = 4
Sample mean, x' = 20
Level of significance, a = 0.98 = 1 - 0.98 = 0.02
The degrees of freedom, df, for a t-distribution = n - 1 = 25 - 1 = 24
Using the t table, the Critical value = [tex] t_\alpha _/_2, _d_f = t_0_._0_2_/_2, _2_4 = t_0_._0_1_, _2_4 = 3.4668 [/tex]
Margin of error, E = [tex] t_\alpha _/_2, _d_f * \frac{\sigma}{\sqrt{n}} = 3.4668 * \frac{4}{\sqrt{25}} = 2.773 [/tex]
Limits of 98% confidence interval, we have:
Lower limit : x' - M.E = 20 - 2.773 = 17.227
Upper limit: x' + M.E = 20 + 2.773 = 22.773
Therefore, (17.227, 22.773) is 98% confidence interval.
b) Let's the margin of error by taking half the length of the confidence interval.
Since we are to use half the length of CI, we have:
[tex] M.E = \frac{22.773 - 17.227}{2} = 2.773 [/tex]
c)[tex] M.E = t_\alpha _/_2, _d_f *s/\sqrt{n} [/tex]
[tex] = t * \frac{s}{\sqrt{n}} = 3.4668 * \frac{4}{\sqrt{25}} = 2.773 [/tex]
