Brainiest to whoever right plz
1 answer:
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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 =
Margin of error, E =
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:
c)