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In the field of statistics, the confidence interval formula refers to the uncertainty associated with a sample population estimate of an actual population.
- It is the representation of the frequency of the confidence intervals that consist of the actual value of the unknown parameter.
- The formula determines the value of the parameter being estimated.
- The value of the interval is based on the standard normal distribution, where the value of Z is the z-score.
- It is also known as the degree of confidence or confidence coefficient.
- Factors in the confidence interval formula include the sample size, the variability, and the confidence level.
For instance, if you establish a confidence interval with a 97 per cent confidence level, it means that we would expect 97% of the interval estimates to include the population parameter.
Key Terms: Statistics, Confidence Level, Confidence Interval Formula, Mean, Standard Normal Distribution, Sample Size, Variability, Degree of Confidence, Range, Population, Probability
What is Confidence Interval Formula?
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A confidence interval, or CI, in statistics, is another way of describing the probability of a parameter. It is a measure of the degree of uncertainty associated with a given statistic. The majority of the time, the intervals are used in conjunction with a margin of error.
- The confidence interval is calculated using an endless number of independent samples.
- Confidence level, which is expressed in percentage, is usually selected by the investigator.
- The mean of the data plus and minus the variation in that data is what gives the confidence interval.
- Inside a specific degree of confidence, this is the range of values you expect your estimate to fall within if you redo the test.
- In most of the examples, the confidence level chosen is 95%.
The alpha value determines the confidence level, which is the percentage of times you expect to replicate an estimate between the top and lower boundaries of the confidence interval.
Real Life Application of Confidence Interval FormulaExample: Confidence Interval Formula is used in the field of biology to determine the mean weight, height, width and diameter of different plant and animal species. |
Confidence Interval Formula
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Confidence Interval Formulas
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A confidence interval formula specifies the range of possibilities for the real value of a parameter. The investigator chooses the confidence level (in percentage). The confidence interval widens as the confidence level rises (less precise).
- It is important to understand and get familiarized with basic statistics formulae.
- The z-score calculation needs to be studied before studying the confidence interval.
- The formula deals with the uncertainty associated with a sampling method.
- The confidence interval formulas for two cases are as follows:
Case 1
In case the value of n is greater than or equal to 30, that is, n ≥ 30, then,
Confidence Interval = \(\overline {x}\) ± zc(σ/√n)
Case 2
In case the value of n is less than 30, that is, n < 30, then,
Confidence Interval = \(\overline {x}\) ± tc(S/√n)
Where,
- n = Number of terms
- \(\overline {x}\) = Sample Mean
- σ = Standard Deviation
- zc = Value corresponding in z -table
- tc = Value corresponding in t- table
- S = Sample standard deviation
The table below showcase the confidence interval formula:
| Confidence Interval Formulas | |
|---|---|
| If n ≥ 30 | Confidence Interval = x̄ ± zα/2(σ/√n) |
| If n<30 | Confidence Interval = x̄ ± tα/2(S/√n) |
Example of Confidence Interval FormulasExample: An automotive engineer wants to estimate the cost of repairing a car that experiences a 25 MPH head-on collision. He crashes 24 cars, and the average repair is $21,000. The standard deviation of the 24-car sample is $1,500. Provide a 98% confidence interval for the true mean cost of repair. Ans: Standard deviation for the sample mean: 1500/√24 =306.74 Since n < 30, we must use the t-table (not the z-table). The 98% t-value for n=24 is 2.5. 21,000±2.5×306.74=21,000±766.87 |
How to Calculate Confidence Interval?
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The steps to calculate confidence interval are as follows:
- Determine a population and then choose a representative sample from it.
- Make a note of the sample number (n).
- Add all of the sample values together and divide by n to get the required mean value.
- Choose a figure for CL (usually 95 percent or more) and look it up in the t-table or z-table as suitable.
- Calculate the standard deviations by subtracting each value from the mean value in the sample.
- After that, square each result and find the average of all the squared differences.
- This will be the variance required.
- Calculate the variance's square root.
- Finally, using the formula mentioned above, determine the CI.
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Things to Remember
- A confidence interval formula is used in statistics to measure the degree of uncertainty associated with a given parameter.
- It will tell us how much confidence we have in the calculated range.
- Usually, 95 percent is taken as confidence level while calculating CI.
- It is used to determine various population parameters for plants, animals, and other life.
- The confidence interval formula is used in the fields of business, finance, manufacturing, market research, medicine, and population studies.
- The mean of the data plus and minus the variation in that data is what gives the confidence interval.
Sample Questions
Ques. A random sample of 30 apples was taken from a large population. On measuring their diameter the mean diameter of the sample was 91 mm with a standard deviation of 8 mm. Calculate the 85% confidence limits for the mean diameter of the whole population of apples. (4 marks)
Ans: For 85% confidence, Z = 1.440 We have: x¯=91,s=8,Z=1.440 and n=30
Substitute into confidence interval formula:
¯x±Zs√nx±Zsn
Therefore, the 85% confidence limits are:
=91±1.440×8√30
=91±1.440×85.47
=91±2.1=91±2.1
The 85% confidence limits are =91±2.1
Ques. Determine the average age of victims of chain snatching in any city during last year for a sample of size 100 with mean age 34.25 years. (3 marks)
Ans: n = 100.
Mean = 34.25 years.
Now, use the 95% CL here. It has z value of 1.96.
Now, calculate the standard deviation. With a mean value of 34.25 and a standard deviation value of 10,a margin of error will be
CI=X¯±z(σ/√n)
CI=34.25±1.96(101/√100)
Thus, Upper limit of CI will be 34.25 + 1.96 = 36.21
And Lower limit of CI will be 34.25 – 1.96 = 32.29
Therefore, we can say with 95% confidence, average age will fall between 32.29 and
36.21 years.
Ques. A random sample of ten scores obtained by the students in a Mathematics test are as follows: 2, 16,3,10,11,4,6,7,9,12. What will be the 90% confidence limits for the mean of the whole sample. (4 marks)
Ans: First, calculate the sample mean:
¯x=(2+3+4+6+7+9+10+11+12+16 )/10
=80/10=8
Now the sample standard deviation:
Since this is a sample, we use the formula s=√1n-1∑ni=1(xi-¯x)2s=1n-1∑i=1n(xi-x)2
with ¯x=8 and n-1=9
On solving for s,
√(19((2-8)2+(3-8)2+(4-8)2+(6-8)2+(7-8)2+(9-8)2+(10-8)2+(11-8)2+(12-8)2+(16-8)2))=√(19(36+25+16+4+1+1+4+9+16+64))
=√(1/9×176)
=√19.555
=4.4221
So for 90% confidence, Z=1.645
Hence, the 90% confidence limits are ¯x±Zs√nx±Zsn
=8±1.645×4.4221/√10
=8±1.645×4.4221/3.1622
=8±2.3
The 90% confidence limits are 8±2.3
Ques. Out of hundreds of people. You randomly chose 46 men with a mean of 86 inches (height) with a standard deviation of 6.2 inches. Determine that the selected men are tall enough. (3 marks)
Ans: Given: Mean, \(\overline {x}\) = 86
Standard deviation, σ = 6.2
Total observations, n = 46
Considering the confidence level to be 95%, the value of z = 1.960
The formula for CI is
\(\overline {x}\) ± Zα/2 × [ σ / √n ]
On substituting the values, we get
86 ± 1.960 × [ 6.2 / √46 ]
86 ± 1.960 × [ 6.2 / 6.78]
86 ± 1.960 × 0.914
86 ± 1.79
So, the margin of error is 1.79
Therefore, all the people are likely to be in the range of 84.21 and 87.79 inches.
Ques. A field agent needs to visit medical clients on a daily basis to conduct health assessments. The agent gets a weekly list of 60 clients to visit during a regular 5 day work week. She has to complete a minimum of 10 visits a day. She completes 3 surveys a day. What is the 95% CI to complete her clients' list by the end of the week. (5 marks)
Ans: Given, one agent requires 60 clients to visit in 5 regular working days. So, the agent needs to complete a minimum of 10 visits a day.
So, for 3 visits a day. The agent completes 3×5=15 visits in a week out of a total number of 60 visits.
The sample proportion of visits the agent completes in a week is 15/60=0.25
Therefore, the sample proportion is ^p=0.25.
Consider, 95% confidence interval for the proportion.
Let the size of the sample be the number of visits that agent needs to complete by the end of the week. Therefore, n=60
The Z-value with 0.05 level of significance is obtained as 1.96 from the standard normal table. Thus, the 95% CI for the proportion of visits completed by the agent is computed as follows:
=0.25±1.96√0.25(1−0.25)60
=0.25±1.96×0.0559
=(0.25±0.11)
=(0.14,0.36)
Therefore, the 95% CI that the agent completes the list of clients by the end of the week is
(0.14×60,0.36×60)=(8.4,21.6)≈(8,22)
Ques. An automotive engineer wants to estimate the cost of repairing a car that experiences a 25 MPH head-on collision. He crashes 24 cars, and the average repair is $11,000. The standard deviation of the 24-car sample is $2,500. Provide a 98% confidence interval for the true mean cost of repair. (3 marks)
Ans: Standard deviation for the sample mean:
2500/√24
=510.2
Since n < 30, we must use the t-table (not the z-table).
The 98% t-value for n=24 is 2.5.
11,000±2.5×510.2=11,000±1275.5
Ques. 300 hundred eggs were randomly chosen from a gravid female salmon and individually weighed. The mean weight was 0.978 g with a standard deviation of 0.042. Find the 95% confidence interval for the mean weight of the salmon eggs (because it is a large n, use the standard normal distribution). (4 marks)
Ans: Since there is such a large sample size, we are using the standard normal or z-distribution in order to calculate the confidence interval.
x¯±z⋅σ/√n
x¯=0.978g
σ=0.042
We must find the appropriate z-value based on the given α for 95% confidence:
1−α2=1−(0.052)
=0.975
Then, find the associated z-score using the z-table for
z0.975=1.96
Now we fill in the formula with our values from the problem to find the 95% CI.
0.978±1.96(0.042÷√300)
=(0.973g,0.9827g)
| Subject | Horn Length (in) | Subject | Horn Length (in) |
|---|---|---|---|
| 1 | 19.1 | 11 | 11.6 |
| 2 | 14.7 | 12 | 18.5 |
| 3 | 10.2 | 13 | 28.7 |
| 4 | 16.1 | 14 | 15.3 |
| 5 | 13.9 | 15 | 13.5 |
| 6 | 12.0 | 16 | 7.7 |
| 7 | 20.7 | 17 | 17.2 |
| 8 | 8.6 | 18 | 19.0 |
| 9 | 24.2 | 19 | 20.9 |
| 10 | 17.3 | 20 | 21.3 |
Ques. The data above represents measurements of the horn lengths of African water buffalo that were raised on calcium supplements. Construct a 95% confidence interval for the population mean for horn length after supplements. (4 marks)
Ans: Calculate the sample mean and the sample standard deviation of the sample.
x¯=16.53
σ=5.29
Since, we do not know the population standard deviation we will using the t-distribution to calculate the confidence intervals.
x¯±t⋅ s/√n
T-value for 95% confidence interval:
α=0.05,0.052=0.025
df=n−1=20−1=19
Look up tα2,19 in t-table and the corresponding t-value = 2.093.
Thus the 95% confidence interval is:
16.53±2.093(5.29/√19)
= (13.99mmHg,19.07mmHg)
Ques. A sample of n=56 observations of 02 consumption by adult western fence lizards gave the following statistics:Mean=249mm3/g/hrs=24mm3/g/hr. Find the 90% confidence limit for the mean Oxygen consumption by adult western fence lizards. ( 3 marks)
Ans: Given: sample standard deviation so, we will use the t-distribution to calculate the confidence interval.
=x¯±ts/√n
Now we must identify our variables:
x¯=249
s=24
n=56
Find the t-value based on the given α
t-value at 90% confidence:
tα2,n−1=t0.102=t0.05,55
Look up t-value for 0.05, 55, so t-value= ~ 1.6735
90% CI becomes:
249±1.6735(24/√56)
=(243.63mm3/g/hr,254.36mm3/g/hr)
Ques. A random sample of 40 apples was taken from a large population. On measuring their diameter the mean diameter of the sample was 90 mm with a standard deviation of 8 mm. Calculate the 85% confidence limits for the mean diameter of the whole population of apples. (4 marks)
Ans: For 85% confidence, Z = 1.440 We have: x¯=01,s=8,Z=1.440 and n=40
Substitute into confidence interval formula:
¯x±Zs√nx±Zsn
Therefore, the 85% confidence limits are:
=90±1.440×8√40
=90±1.440×50.59
=90±3.1=91±3.1
The 85% confidence limits are =90±3.1
Ques. An automotive engineer wants to estimate the cost of repairing a car that experiences a 25 MPH head-on collision. He crashes 24 cars, and the average repair is $33,000. The standard deviation of the 24-car sample is $1,600. Provide a 98% confidence interval for the true mean cost of repair. (3 marks)
Ans: Standard deviation for the sample mean:
1600/√24
=327.19
Since n < 30, we must use the t-table (not the z-table).
The 98% t-value for n=24 is 2.5.
33,000±2.5×327.19=33,000±817.9
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