What is a 90% confidence interval for the mean and what sample size is required

Question

What is a 90% confidence interval for the mean and what sample size is required to estimate this within 1 unit?

51.93965 53.8713 46.61498 53.88235 58.68809 52.49902 55.19509 48.61611 61.13262 46.85473 50.94638 48.60558 33.87669 48.0081 58.71772 45.0263 51.3847 43.4046 67.13301 50.25178 51.90914 43.241 48.39393 49.94075 49.50116 49.03525 56.16066 54.18414 41.40659 46.17212 48.59269 49.42387 51.2602 51.89276 50.39931 42.07651 49.68158 58.70999 64.30357 46.62885 51.62401 57.79418 50.26903 57.55324 51.63657 56.76422 50.70841 44.57391 49.39527 53.51338

Explanation / Answer

51.93965,53.8713,46.61498,53.88235,58.68809,52.49902,55.19509,48.61611,61.13262,46.85473,50.94638,48.60558,33.87669,48.0081,58.71772,45.0263,51.3847,43.4046

Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=50.5147
Standard deviation( sd )=6.4072
Sample Size(n)=18
Confidence Interval = [ 50.5147 ± t a/2 ( 6.4072/ Sqrt ( 18) ) ]
= [ 50.5147 - 1.74 * (1.51) , 50.5147 + 1.74 * (1.51) ]
= [ 47.887,53.142 ]

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.