Confidence Interval for the Mean Calculator

Enter N	Enter X	Enter σ or s	Enter Confidence Interval %	Rounding Digits

   

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A small sample of 10 units has a mean μ and a standard deviation σ of 1

Find a 98% confidence interval of the mean μ

Confidence Interval Formula for μ

X - tscore_α * s/√n < μ < X + tscore_α * s/√n where:

X = sample mean, s = sample standard deviation, tscore = statistic with (n - 1) Degrees of Freedom and α = 1 - confidence Percentage

Find degrees of freedom:

Degrees of Freedom = n - 1

Degrees of Freedom = 10 - 1

Degrees of Freedom = 9

Calculate α

α = 1 - Confidence%
α = 1 - 0.98
α = 0.02

Find α spread range:

α = ½(α)

α = ½(0.02)

α = 0.01

Calculate t-score

Find t-score for α_0.01 using 9 degrees of freedom:

tscore_0.01 = 2.8214 <--- Value can be found on Excel using =TINV(0.02,9)

tscore = 2.8214

Locate Value in the t-chart below:

_DOF^αα _0.005α _0.01α _0.015α _0.02α _0.025α _0.03α _0.035α _0.04α _0.045α _0.05163.655931.82121.205115.894512.706210.57899.05797.91587.02646.313729.9256.96455.64284.84874.30273.89643.57823.31983.1042.9235.84084.54073.89613.48193.18242.95052.76262.60542.47082.353444.60413.74693.29762.99852.77652.60082.45592.33292.22612.131854.03213.36493.00292.75652.57062.42162.29742.1912.09782.01563.70743.14272.82892.61222.44692.31332.20112.10432.01921.943273.49952.99792.71462.51682.36462.24092.13652.0461.96621.894683.35542.89652.63382.4492.3062.18922.09022.00421.9281.859593.24982.82142.57382.39842.26222.15042.05541.97271.89921.8331103.16932.76382.52752.35932.22812.12022.02831.94811.87681.8125113.10582.71812.49072.32812.2012.09612.00671.92841.85881.7959123.05452.6812.46072.30272.17882.07641.98891.91231.8441.7823133.01232.65032.43582.28162.16042.061.97421.89891.83171.7709142.97682.62452.41492.26382.14482.04621.96171.88751.82131.7613152.94672.60252.3972.24852.13152.03431.95091.87771.81231.7531162.92082.58352.38152.23542.11992.0241.94171.86931.80461.7459172.89822.56692.36812.22382.10982.0151.93351.86191.79781.7396182.87842.55242.35622.21372.10092.00711.92641.85531.79181.7341192.86092.53952.34572.20472.09321.921.84951.78641.7291202.84532.5282.33622.19672.0861.99371.91431.84431.78161.7247212.83142.51762.32782.18942.07961.9881.90921.83971.77731.7207222.81882.50832.32022.18292.07391.98291.90451.83541.77341.7171232.80732.49992.31322.1772.06871.97831.90031.83161.76991.7139242.7972.49222.30692.17152.06391.9741.89651.82811.76671.7109252.78742.48512.30112.16662.05951.97011.89291.82481.76371.7081262.77872.47862.29582.1622.05551.96651.88971.82191.7611.7056272.77072.47272.29092.15782.05181.96321.88671.81911.75851.7033282.76332.46712.28642.15392.04841.96011.88391.81661.75611.7011292.75642.4622.28222.15032.04521.95731.88131.81421.7541.6991

Calculate the Standard Error of the Mean:

SEM  =	σ
	√n

SEM  =	1
	√10

SEM  =	1
	3.1622776601684

SEM = 0.3162

Calculate high end confidence interval total:

High End = X + tscore_α * s/√n

High End = 5 + 2.8214 x 1/√10

High End = 5 + 2.8214 x 0.31622776601684

High End = 5 + 0.89220501903991

High End = 5.8922

Calculate low end confidence interval total:

Low End = X - tscore_α * s/√n

Low End = 5 - 2.8214 x 1/√10

Low End = 5 - 2.8214 x 0.31622776601684

Low End = 5

Final Answer

What this means is if we repeated experiments, the proportion of such intervals that contain μ would be 98%