Let a sample have the following characteristics :
- Standard deviation : \sigma=4
- Mean : \bar{x}=36
We need to have a 95% probability for the population mean to be between 34 and 38.
What is the minimum value of n ?
Let z^* be the z -score. The margin of error is:
z^*\dfrac{\sigma}{\sqrt{n}}
We have:
\bar{x}-34 = 38 - \bar{x} = 2
Therefore:
z^*\dfrac{\sigma}{\sqrt{n}} \le 2
According to the z -table, the value of z^* corresponding to the confidence level of 95% is 1.96. So we can write:
1.96 \times \dfrac{4}{\sqrt{n}} \le 2
\sqrt{n} \ge 3.92
n \ge 15.4
The sample size must be greater than or equal to 16.
We weigh a sample of apples of a garden. The mean of the weight of these gardens is 86 grams and the standard deviation is 6.2 grams. We need to have a 95% probability for the mean to be between 84 and 88.
What is the minimum value of n ?
Let z^* be the z -score. The margin of error is:
z^*\dfrac{\sigma}{\sqrt{n}}
We have:
\bar{x}-84 = 88 - \bar{x} = 2
Therefore:
z^*\dfrac{\sigma}{\sqrt{n}} \le 2
According to the z -table, the value of z^* corresponding to the confidence level of 95% is 1.96. So, we can write:
1.96 \times \dfrac{6.2}{\sqrt{n}} \le 2
\sqrt{n} \ge 6.08
n \ge 36.97
The sample size must be greater than or equal to 37.
We choose a sample of bottled water to determine the pH (the acidity scale) of the water produced by a company. The mean of the pH of the sample is 6.52 and the standard deviation is 0.11.
What is the minimum value of n to have a 99% probability for the population mean to be between 6.50 and 6.54 ?
Let z^* be the z -score. The margin of error is:
z^*\dfrac{\sigma}{\sqrt{n}}
We have:
\bar{x}-6.32 = 6.72 - \bar{x} = 0.02
Therefore:
z^*\dfrac{\sigma}{\sqrt{n}} \le 0.02
According to the z -table, the value of z^* corresponding to the confidence level of 99% is 2.58. So, we can write:
2.58 \times \dfrac{0.11}{\sqrt{n}} \le 0.02
\sqrt{n} \ge 14.19
n \ge 201.36
The sample size must be greater than or equal to 202.
We measure the height of a sample of people. The mean of the sample is 175 cm and the standard deviation is 20 cm.
What is the minimum value of n to have a 95% probability for the population mean to be between 170 and 180 ?
Let z^* be the z -score. The margin of error is:
z^*\dfrac{\sigma}{\sqrt{n}}
We have:
\bar{x}-170 = 180 - \bar{x} = 5
Therefore:
z^*\dfrac{\sigma}{\sqrt{n}} \le 5
According to the z -table, the value of z^* corresponding to the confidence level of 95% is 1.96. So, we can write:
1.96 \times \dfrac{20}{\sqrt{n}} \le 5
\sqrt{n} \ge 7.84
n \ge 61.47
The sample size must be greater than or equal to 62.
We choose a sample of the grades of n students in a math class. The mean of the grades is 14 and the standard deviation is 4.
What is the minimum value of n to have a 90% probability for the mean to be between 12.5 and 15.5 ?
Let z^* be the z -score. The margin of error is:
z^*\dfrac{\sigma}{\sqrt{n}}
We have:
\bar{x}-12.5 = 15 - \bar{x} = 1.5
Therefore:
z^*\dfrac{\sigma}{\sqrt{n}} \le 1.5
According to the z -table, the value of z^* corresponding to the confidence level of 90% is 1.65. So, we can write:
1.65 \times \dfrac{4}{\sqrt{n}} \le 1.5
\sqrt{n} \ge 4.4
n \ge 19.36
The sample size must be greater than or equal to 20.
We choose a sample n bottles of juice. The mean of the volume of sample is 997 ml and the standard deviation is 15 ml.
What is the minimum value of n to have a 99% probability for the population mean to be between 990 ml and 1\ 004. ml?
Let z^* be the z -score. The margin of error is:
z^*\dfrac{\sigma}{\sqrt{n}}
We have:
\bar{x}-990 = 1\ 004 - \bar{x} = 7
Therefore:
z^*\dfrac{\sigma}{\sqrt{n}} \le 7
According to the z -table, the value of z^* corresponding to the confidence level of 99% is 2.58. So, we can write:
2.58 \times \dfrac{15}{\sqrt{n}} \le 7
\sqrt{n} \ge 5.53
n \ge 30.57
The sample size must be greater than or equal to 31.
A medical center chooses a sample of n clients who regularly smoke. Assume that the mean of the ages of the clients is 43 and the standard deviation is 12.
What is the minimum value of n to have a 90% probability for the population mean to be between 37 and 49.
Let z^* be the z -score. The margin of error is:
z^*\dfrac{\sigma}{\sqrt{n}}
We have:
\bar{x}-37 = 49 - \bar{x} = 6
Therefore:
z^*\dfrac{\sigma}{\sqrt{n}} \le 6
According to the z -table, the value of z^* corresponding to the confidence level of 90% is 1.65. So, we can write:
1.65 \times \dfrac{12}{\sqrt{n}} \le 6
\sqrt{n} \ge 3.3
n \ge 10.89
The sample size must be greater than or equal to 11.