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  1. Home
  2. 12th grade
  3. Statistics & Probabilities
  4. Exercise : Find confidence intervals for population means and/or population proportion when the sample size, the standard deviation and the sample mean are given

Find confidence intervals for population means and/or population proportion when the sample size, the standard deviation and the sample mean are given Statistics & Probabilities

A sample has the following characteristics :

  • Size : n=64
  • Standard deviation : \sigma=2
  • Mean : \bar{x}=13

Find a confidence interval for the population mean with a confidence of 95%.

The confidence interval when the standard deviation is known can be computed using the following formula:

\left[ \overline{x}-z^*\dfrac{\sigma}{\sqrt{n}},\overline{x}+z^*\dfrac{\sigma}{\sqrt{n}} \right]

According to the z -table, the value of z^* corresponding to confidence of 95 percent is 1.96. So we have:

z^*\dfrac{\sigma}{\sqrt{n}} = 1.96\dfrac{2}{\sqrt{64}}=0.49

The confidence interval is:

\left[ 13- 0.49 , 13 + 0.49\right]

Or:

\left[ 12.51, 13.49 \right]

The population mean has a 95% chance to be on \left[12.51{,}13.49\right]

A sample has the following characteristics :

  • Size : n=100
  • Standard deviation : \sigma=18
  • Mean : \bar{x}=86

Find a confidence interval for the population mean with a confidence of 95%.

The confidence interval when the standard deviation is known can be computed using the following formula:

\left[ \overline{x}-z^*\dfrac{\sigma}{\sqrt{n}},\overline{x}+z^*\dfrac{\sigma}{\sqrt{n}} \right]

According to the z -table, the value of z^* corresponding to confidence of 95 percent is 1.96. So we have:

z^*\dfrac{\sigma}{\sqrt{n}} = 1.96\dfrac{18}{\sqrt{100}}=3.53

The confidence interval is:

\left[ 86-3.53, 86+3.53\right]

Or:

\left[ 82.47{,}89.53 \right]

The population mean has a 95% chance to be on \left[ 82.47{,}89.53 \right]

A sample has the following characteristics :

  • Size : n=625
  • Standard deviation : \sigma=50
  • Mean : \bar{x}=145

Find a confidence interval for the population mean with a confidence of 99%.

The confidence interval when the standard deviation is known can be computed using the following formula:

\left[ \overline{x}-z^*\dfrac{\sigma}{\sqrt{n}},\overline{x}+z^*\dfrac{\sigma}{\sqrt{n}} \right]

According to the z -table, the value of z^* corresponding to confidence of 99 percent is 2.58. So we have:

z^*\dfrac{\sigma}{\sqrt{n}} = 2.58\dfrac{50}{\sqrt{625}}= 5.16

The confidence interval is:

\left[ 145-5.16, 145+5.16\right]

Or:

\left[ 139.84{,}150.16 \right]

The population mean has a 99% chance to be on \left[ 139.84{,}150.16 \right]

A sample has the following characteristics :

  • Size : n=289
  • Standard deviation : \sigma=68
  • Mean : \bar{x}=401

Find a confidence interval for the population mean with a confidence of 95%.

The confidence interval when the standard deviation is known can be computed using the following formula:

\left[ \overline{x}-z^*\dfrac{\sigma}{\sqrt{n}},\overline{x}+z^*\dfrac{\sigma}{\sqrt{n}} \right]

According to the z -table, the value of z^* corresponding to confidence of 95 percent is 1.96. So we have:

z^*\dfrac{\sigma}{\sqrt{n}} = 1.96\dfrac{68}{\sqrt{289}}=7.84

The confidence interval is:

\left[ 401-7.84, 401+7.84\right]

Or:

\left[ 393.16{,}408.84 \right]

The population mean has a 95% chance to be on \left[ 393.16{,}408.84 \right]

A sample has the following characteristics :

  • Size : n=324
  • Standard deviation : \sigma=18
  • Mean : \bar{x}=99

Find a confidence interval for the population mean with a confidence of 95%.

The confidence interval when the standard deviation is known can be computed using the following formula:

\left[ \overline{x}-z^*\dfrac{\sigma}{\sqrt{n}},\overline{x}+z^*\dfrac{\sigma}{\sqrt{n}} \right]

According to the z -table, the value of z^* corresponding to confidence of 95 percent is 1.96. So we have:

z^*\dfrac{\sigma}{\sqrt{n}} = 1.96\dfrac{18}{\sqrt{324}}=1.96

The confidence interval is:

\left[ 99-1.96{,}99+1.96\right]

Or:

\left[97.04, 100.96 \right]

The population mean has a 95% chance to be on \left[97.04, 100.96 \right]

A sample has the following characteristics :

  • Size : n=81
  • Standard deviation : \sigma= 27
  • Mean : \bar{x}=55

Find a confidence interval for the population mean with a confidence of 90%.

The confidence interval when the standard deviation is known can be computed using the following formula:

\left[ \overline{x}-z^*\dfrac{\sigma}{\sqrt{n}},\overline{x}+z^*\dfrac{\sigma}{\sqrt{n}} \right]

According to the z -table, the value of z^* corresponding to confidence of 90 percent is 1.65. So we have:

z^*\dfrac{\sigma}{\sqrt{n}} = 1.65\dfrac{27}{\sqrt{81}}=4.95

The confidence interval is:

\left[ 55-4.95, 55+4.95\right]

Or:

\left[ 50.05{,}59.95 \right]

The population mean has a 90% chance to be on \left[ 50.05{,}59.95 \right]

A sample has the following characteristics :

  • Size : n=256
  • Standard deviation : \sigma=32
  • Mean : \bar{x}=75

Find a confidence interval for the population mean with a confidence of 99%.

The confidence interval when the standard deviation is known can be computed using the following formula:

\left[ \overline{x}-z^*\dfrac{\sigma}{\sqrt{n}},\overline{x}+z^*\dfrac{\sigma}{\sqrt{n}} \right]

According to the z -table, the value of z^* corresponding to confidence of 99 percent is 2.58. So we have:

z^*\dfrac{\sigma}{\sqrt{n}} = 2.58\dfrac{32}{\sqrt{256}}=5.16

The confidence interval is:

\left[ 75-5.16, 75+5.16\right]

Or:

\left[ 69.84{,}80.16 \right]

The population mean has a 99% chance to be on \left[ 69.84{,}80.16 \right]

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See also
  • Course : Hypothesis testing and estimation
  • Exercise : Use the central limit theorem to find the probability that the sample mean is between two given real numbers
  • Exercise : Determine whether samples are biased or unbiased
  • Exercise : Determine the sample size needed for a level of confidence
  • Exercise : Identify whether an experiment is well constructed
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