A confidence interval is a statistical formula that indicates sample measurement’s precisions, it also predicts how close a sample data is to the original estimate (Greenland et al., 2016). In addition, the function describes the uncertainties linked to a sampling method used. One needs a confidence level, such as 99%, 95%, or 90% to calculate a confidence interval. A 95% confidence level implies that there is only a 5 % chance of interval estimate to be out of population parameter.

A confidence interval is also defined as a value range that one is sure the exact value lies in. There are two formulas used in calculating it based on the sample size. Z score is used for more than 30 samples while t- table is used when the size is less than 30 (Greenland et al., 2016). Therefore, if n ≥ 30, the confidence interval is x̄ ± z_{α/2}(σ/√n) while when n<30 it is x̄ ± t_{α/2}(S/√n). Where,

n = Number of terms

x̄ = Sample Mean

σ = Standard Deviation

z_{α/2} = Value corresponding to α2 in z table

t_{α/2} = Value corresponding to α2 in t table

α = (1 – Confidence Level /100)

## Sample question

The height of students from a class was sampled randomly and measurements were recorded as follows:

95, 108, 97, 112, 99, 106, 105, 100, 99, 98, 104, 110, 107, 111, 103, 110.

Calculate their confidence interval at 95%.

## Solution

Confidence Interval = x̄ ± t_{α/2}(S/√n)

Df = n−1, which is 16-1=15

Sample mean = (95+ 108+ 97+ 112+ 99+106 + 105 +100 + 99 +98+ 104 +110 +107 + 111 + 103 + 110)/16

x̄ =104

α= (1− decimal confidence level)/2=

=2(1−0.95) =0.025

95 z α/2=1.96

CI _{= }(104-1.96 ×1.25, 104+1.9× 1.25) = (101.55; 106.45)

### Response One

A poll to ascertain how many dogs are owned by seven different members conducted and the results breakdown is as follows 2, 3, 0, 1, 1, 2, and 2.

The formulae for interval calculation will not be Z score because the n value in this case is less than 30. Therefore, t-tables will be utilized to check the statistics and the formulae to be used in calculating confidence interval (Greenland et al., 2016). This step is critical to minimize the margin that occurs with smaller sample size. T values depends on degree of freedom, which makes the probability value to be limited because n is usually the denominator. Utilization of a 95 % confidence level for this question implies that the confidence interval will have 0.05 chance of not containing the true population mean.

Sample mean** = (**2+ 3 + 0 + 1 + 1 + 2+ 2)/7 = 1.5714285

Standard deviation**=**0.95238095

CI= x̄ ± t_{α/2}(S/√n)

α= (1− decimal confidence level) /2 = (1−0.95)/2=0.025

n=7, df =7-1=6

T- Value = (6, 0.025) = 2.447

The corresponding two tailed T – value from t-tables for 95% confidence level at a degree level of 6 is 2.447. Thus, the confidence interval of lower and upper limit is mean ± 2.447 ×standard. Thus,

=1.5714285± t×0.95238095/√7

=1.5714285± (2.447 ×0 95238095/√7)

=1.5714285± 0.880837203

= (1.5714285 + 0.880837203), (1.5714285 – 0.880837203),

= (2.4522657, 0.690591297)

The number of dogs in my friends or family houses is either 2, 1, or and 0 when 95% confidence interval is utilized.

### Response Two

The total number of sold vehicles is 432 and the SUVs variety is 142.Thus,

P=142/432=0.33

The confidence level given above is 90%

Z=1.645

So, the confidence level=1.645

Thus, the confidence interval would be level

0.00372

(0.33+0.00372) (0.33-0.00372)

(0.2928, 0.3672)

This implies that there is a 33 % chance of a vehicle being sold to be an SUV or rather in a group of 100 about 33 SUVs will be sold. The sample error for this population should be about 4.

Since the sample size is greater than 30, it is correct to use Z score (Greenland et al., 2016).

The confidence interval formulae is supposed to be: the probability ± 1/ √n. Thus, the formula shown below is correct

Since the data given follows a normal distribution and it has a larger sample size of more than 30, it was appropriate for the student to use a z distribution in calculating the required critical values. In addition, the z statistics, which is1.645 given corresponds to a 90% confidence level in Z score table.

Test done on larger population sample of more than 30 have a relatively wider confidence interval than those of small sample. This implies that the imprecision at 90 % confidence interval is lower for this question. Thus, there is an assurance that the confidence interval for selling SUVs is between 0.2928 and 0.3672. The result signifies that about 29 to 39 vehicles sold out of 100 will be SUVs.

## Reference

Greenland, S., Senn, S. J., Rothman, K. J., Carlin, J. B., Poole, C., Goodman, S. N., & Altman, D. G. (2016). Statistical tests, P values, confidence intervals, and power: A guide to misinterpretations. *European Journal of Epidemiology*, *31*(4), 337-350. Web.