Confidence Interval worksheet

1. 99% confidence interval: 0.3617 to 0.5716

1. Find your sample's share (p̂): the successes divided by the sample size

   $\mathrm{p̂}$$=\frac{70}{150}$$=0.466667$

2. Look up the critical z value for 99% confidence. It sets how many standard errors the interval reaches on each side

   $\mathrm{z*}$$=2.5758$

3. Find the standard error: how much your sample's share would vary from one sample to the next

   $\mathrm{SE}$$=\sqrt{\frac{0.466667\times \left(1-0.466667\right)}{150}}$$=0.040734$

4. Multiply the critical value by the standard error to get the margin of error, how far the interval reaches each way

   $E$$=2.5758\times 0.040734$$=0.104924$

5. Subtract and add the margin to your sample's share to get the interval

   $\mathrm{CI}$$=0.466667\text{±}0.104924$$=\text{0.3617 to 0.5716}$

2. 95% confidence interval: 35.5662 to 44.4338

1. The standard deviation comes from the sample, so use the t distribution with n − 1 = 21 degrees of freedom. Its critical value for 95% confidence is

   $\mathrm{t*}$$=2.0796$

2. Find the standard error: the standard deviation divided by the square root of the sample size. It shows how much the average would vary between samples

   $\mathrm{SE}$$=\frac{10}{\sqrt{22}}$$=2.132007$

3. Multiply the critical value by the standard error to get the margin of error, how far the interval reaches each way

   $E$$=2.0796\times 2.132007$$=4.433752$

4. Subtract and add the margin to the sample mean to get the interval

   $\mathrm{CI}$$=40\text{±}4.433752$$=\text{35.5662 to 44.4338}$

3. 99% confidence interval: 42.1526 to 57.8474

1. The standard deviation comes from the sample, so use the t distribution with n − 1 = 21 degrees of freedom. Its critical value for 99% confidence is

   $\mathrm{t*}$$=2.8314$

2. Find the standard error: the standard deviation divided by the square root of the sample size. It shows how much the average would vary between samples

   $\mathrm{SE}$$=\frac{13}{\sqrt{22}}$$=2.771609$

3. Multiply the critical value by the standard error to get the margin of error, how far the interval reaches each way

   $E$$=2.8314\times 2.771609$$=7.847423$

4. Subtract and add the margin to the sample mean to get the interval

   $\mathrm{CI}$$=50\text{±}7.847423$$=\text{42.1526 to 57.8474}$

4. 99% confidence interval: 34.6965 to 47.3035

1. The standard deviation comes from the sample, so use the t distribution with n − 1 = 23 degrees of freedom. Its critical value for 99% confidence is

   $\mathrm{t*}$$=2.8073$

2. Find the standard error: the standard deviation divided by the square root of the sample size. It shows how much the average would vary between samples

   $\mathrm{SE}$$=\frac{11}{\sqrt{24}}$$=2.245366$

3. Multiply the critical value by the standard error to get the margin of error, how far the interval reaches each way

   $E$$=2.8073\times 2.245366$$=6.303495$

4. Subtract and add the margin to the sample mean to get the interval

   $\mathrm{CI}$$=41\text{±}6.303495$$=\text{34.6965 to 47.3035}$

5. 99% confidence interval: 0.6073 to 0.8214

1. Find your sample's share (p̂): the successes divided by the sample size

   $\mathrm{p̂}$$=\frac{80}{110}$$=0.727273$

2. Look up the critical z value for 99% confidence. It sets how many standard errors the interval reaches on each side

   $\mathrm{z*}$$=2.5758$

3. Find the middle of the Wilson interval. It's your sample's share, nudged slightly toward one half

   $\mathrm{center}$$=\frac{0.727273+\frac{{2.5758}^2}{2\times 110}}{1+\frac{{2.5758}^2}{110}}$$=0.714344$

4. Find how far the interval reaches on each side (the half-width). It uses the usual spread of a share plus a small correction

   $h$$=\frac{2.5758}{1+\frac{{2.5758}^2}{110}}\times \sqrt{\frac{0.727273\times \left(1-0.727273\right)}{110}+\frac{{2.5758}^2}{4\times {110}^2}}$$=0.107006$

5. Subtract and add the half-width to get the interval, kept between 0 and 1

   $\mathrm{CI}$$=0.714344\text{±}0.107006$$=\text{0.6073 to 0.8214}$

6. 99% confidence interval: 0.3856 to 0.6144

1. Find your sample's share (p̂): the successes divided by the sample size

   $\mathrm{p̂}$$=\frac{60}{120}$$=0.5$

2. Look up the critical z value for 99% confidence. It sets how many standard errors the interval reaches on each side

   $\mathrm{z*}$$=2.5758$

3. Find the middle of the Wilson interval. It's your sample's share, nudged slightly toward one half

   $\mathrm{center}$$=\frac{0.5+\frac{{2.5758}^2}{2\times 120}}{1+\frac{{2.5758}^2}{120}}$$=0.5$

4. Find how far the interval reaches on each side (the half-width). It uses the usual spread of a share plus a small correction

   $h$$=\frac{2.5758}{1+\frac{{2.5758}^2}{120}}\times \sqrt{\frac{0.5\times \left(1-0.5\right)}{120}+\frac{{2.5758}^2}{4\times {120}^2}}$$=0.114449$

5. Subtract and add the half-width to get the interval, kept between 0 and 1

   $\mathrm{CI}$$=0.5\text{±}0.114449$$=\text{0.3856 to 0.6144}$

7. 99% confidence interval: 39.8426 to 50.1574

1. The standard deviation comes from the sample, so use the t distribution with n − 1 = 23 degrees of freedom. Its critical value for 99% confidence is

   $\mathrm{t*}$$=2.8073$

2. Find the standard error: the standard deviation divided by the square root of the sample size. It shows how much the average would vary between samples

   $\mathrm{SE}$$=\frac{9}{\sqrt{24}}$$=1.837117$

3. Multiply the critical value by the standard error to get the margin of error, how far the interval reaches each way

   $E$$=2.8073\times 1.837117$$=5.157405$

4. Subtract and add the margin to the sample mean to get the interval

   $\mathrm{CI}$$=45\text{±}5.157405$$=\text{39.8426 to 50.1574}$

8. 90% confidence interval: 0.2537 to 0.4009

1. Find your sample's share (p̂): the successes divided by the sample size

   $\mathrm{p̂}$$=\frac{36}{110}$$=0.327273$

2. Look up the critical z value for 90% confidence. It sets how many standard errors the interval reaches on each side

   $\mathrm{z*}$$=1.6449$

3. Find the standard error: how much your sample's share would vary from one sample to the next

   $\mathrm{SE}$$=\sqrt{\frac{0.327273\times \left(1-0.327273\right)}{110}}$$=0.044738$

4. Multiply the critical value by the standard error to get the margin of error, how far the interval reaches each way

   $E$$=1.6449\times 0.044738$$=0.073588$

5. Subtract and add the margin to your sample's share to get the interval

   $\mathrm{CI}$$=0.327273\text{±}0.073588$$=\text{0.2537 to 0.4009}$

9. 95% confidence interval: 72.5406 to 79.4594

1. The population's standard deviation is known, so use the normal (bell-curve) distribution. Its critical value for 95% confidence is

   $\mathrm{z*}$$=1.96$

2. Find the standard error: the standard deviation divided by the square root of the sample size. It shows how much the average would vary between samples

   $\mathrm{SE}$$=\frac{9}{\sqrt{26}}$$=1.765045$

3. Multiply the critical value by the standard error to get the margin of error, how far the interval reaches each way

   $E$$=1.96\times 1.765045$$=3.459425$

4. Subtract and add the margin to the sample mean to get the interval

   $\mathrm{CI}$$=76\text{±}3.459425$$=\text{72.5406 to 79.4594}$

10. 64% confidence interval: 0.2425 to 0.3142

1. Find your sample's share (p̂): the successes divided by the sample size

   $\mathrm{p̂}$$=\frac{36}{130}$$=0.276923$

2. Look up the critical z value for 64% confidence. It sets how many standard errors the interval reaches on each side

   $\mathrm{z*}$$=0.9154$

3. Find the middle of the Wilson interval. It's your sample's share, nudged slightly toward one half

   $\mathrm{center}$$=\frac{0.276923+\frac{{0.9154}^2}{2\times 130}}{1+\frac{{0.9154}^2}{130}}$$=0.278352$

4. Find how far the interval reaches on each side (the half-width). It uses the usual spread of a share plus a small correction

   $h$$=\frac{0.9154}{1+\frac{{0.9154}^2}{130}}\times \sqrt{\frac{0.276923\times \left(1-0.276923\right)}{130}+\frac{{0.9154}^2}{4\times {130}^2}}$$=0.035838$

5. Subtract and add the half-width to get the interval, kept between 0 and 1

   $\mathrm{CI}$$=0.278352\text{±}0.035838$$=\text{0.2425 to 0.3142}$