Confidence Interval: Formula, Table & Calculation

Ans:  Z = 1.440 for 85% of the confidence limits
As per the given equation: \(\overline{\text{X}}\)=91, s=8, Z=1.440 and n=30
Now, we need to substitute the confidence level formula with the values present here,
x ± Zs\(\sqrt{n}\)
Hence, considering 85% confidence limits, the equation will be as follows:
= 91 ± 1.440 × 8\(\sqrt{30}\)
= 91 ± 1.440 × 85.477…
= 91 ± 2.1

Ans: As per the question, the following data can be gathered,
The Sample mean, based on the equation = \(\overline{\text{X}}\)=11.3
Standard deviation = σ=1.5
Sample size = n=40
Confidence level, c, is considered as = 90%
As we are aware of the population standard deviation and the size that is seemingly larger than 30, then we should use the standard normal (z) distribution to attain the confidence interval.
The z-value, for 90% of confidence level, is = 1.645

\(\text{Confidence Interval} = \text{Sample Mean} \pm ( \text{z - value at 0.90} \times \frac{\text{Standard deviation}}{\sqrt{\text{sample size}}})\)

                                \(= 11.3 \pm (1.645 \times \frac{1.5}{\sqrt{40}})\)

                                \(= (10.910 , 11.690)\)

Ans: 
Mean of the whole sample =

 \(\frac{2+3+4+6+7+9+10+11+12+16}{10} =\frac{80}{10} = 8\)
The sample standard deviation:
Since, this is a sample, formulae used is = 1n−1∑ni=1(xi−¯x)2, with ¯x=8 and n−1=9
Therefore s =\(\sqrt{\frac{1}{9}(2-8)^2 + (3-8)^2 +(4-8)^2 + (6-8)^2 + (7-8)^2 + (9-8)^2}\)

                   \(=\sqrt{\frac{1}{9}36+25+16+4+1+1+4+9+16+64}\)  

                   \(= \sqrt{\frac{1}{9} \times 176}\)

                   \(= \sqrt{19.555}\)

                   \(= 4.4221 ..\)

90% confidence (Z) = 1.645
As per the answer, 90% of the confidence limits = 1.645

Ans: The given mean value is = 86
As per the question, Standard Deviation = 6.2 inches
Now, considering the total observations, we get n = 46 men
Therefore, if presuming that the confidence level is 95% = 1.960
Formulae used = \(\overline{\text{X}}\) ± Zα/2 × [ σ / \(\sqrt{n}\) ]
Substituting values, we get:
86 ± 1.960 × [ 6.2 / \(\sqrt{46}\) ]
86 ± 1.960 × [ 6.2 / 6.78]
86 ± 1.960 × 0.914
86 ± 1.79
Hence, the margin of error can be considered as 1.79
Therefore, the selected men are possibly between the height gap of about 84. 21 and 87.79 inches.

Ans: A Confidence Interval (or, CI), in the family of statistics and mathematics, is apparently a broad range of estimates. It can be further broken down into a lower bound and an upper bound for an unknown parameter. It is physically impossible to study every person in the population. It has led researchers to believe in selecting a sample of the whole population as a way to determine the results. In a nutshell, this outlines the idea about researchers being able to estimate the parameters of a population by the provided set of sample data (which clearly is somewhat unknown). Here is the table of confidence interval,

Ans: The confidence level is known to exhibit a frequency of reasonable confidence intervals, which comprises the upper and lower bounds, that includes the true value of the unknown parameter. It can be estimated on the basis of mean and standard deviation. The formula of Confidence Interval (or, CI) is as follows,
\(\overline{\text{X}}\) ± Zα/2 × [ σ / \(\sqrt{n}\)]
Wherein, \(\overline{\text{X}}\) = Mean,
Z = Confidence coefficient,
α = Confidence level,
σ = Standard deviation,
N = sample space,
Whereby, it is significant to know that the value following the “±” symbol is actually referred to as the margin of error.

Ans: It often becomes cumbersome to produce sample confidence intervals for a largely dispersed data, which gradually led researchers to only accept a sample out of the whole to generate accurate results.

• Confidence Interval is usually only accurate if considering normal distribution of the population.
• There are cases that defeat the accuracy. In brief, if the estimation has to be about a large sample from other types of population distributions, then the central limit theorem is utilized for conducting further accuracy of the interval sample set.

Ans. The difference between confidence interval and confidence label are as follows:

Ans:  Z = 1.440 for 85% of the confidence limits
As per the given equation: \(\overline{\text{X}}\)=61, s=8, Z=1.440 and n=25
Now, we need to substitute the confidence level formula with the values present here,
x ± Zs\(\sqrt{n}\)
Hence, considering 85% confidence limits, the equation will be as follows:
= 61 ± 1.440 × 10\(\sqrt{25}\)
= 61 ± 1.440 × 50

Ans: The given mean value is = 96
As per the question, Standard Deviation = 7.2 inches
Now, considering the total observations, we get n = 36 men
Therefore, if presuming that the confidence level is 95% = 1.960
Formulae used = \(\overline{\text{X}}\) ± Zα/2 × [ σ / \(\sqrt{n}\) ]
Substituting values, we get:
96 ± 1.960 × [ 7.2 / \(\sqrt{36}\) ]
96 ± 1.960 × [ 7.2 / 6]
96 ± 1.960 × 1.2
96 ± 2.35
Hence, the margin of error can be considered as 2.35
Therefore, the selected men are possibly between the height gap of about 98.352 and 93.65 inches.

Ans. A range that includes an upper and lower value derived from a sample is known as the 95% confidence interval (CI) of the mean. This range denotes potential mean values because the actual population mean is unknown.