What Happens To The Margin Of Error As The Confidence Level Increases

Introduction

While you are learning statistics, you lot will oft accept to focus on a sample rather than the unabridged population. This is because it is extremely costly, hard and time-consuming to study the entire population. The all-time you lot can do is to take a random sample from the population – a sample that is a 'true' representative of it. You then carry out some analysis using the sample and brand inferences about the population.

Since the inferences are fabricated almost the population by studying the sample taken, the results cannot be entirely accurate. The degree of accurateness depends on the sample taken – how the sample was selected, what the sample size is, and other concerns. Mutual sense would say that if you increase the sample size, the chances of error will be less because yous are taking a greater proportion of the population. A larger sample is likely to be a closer representative of the population than a smaller one.

Let's consider an example. Suppose you want to study the scores obtained in an exam by students in your college. Information technology may be time-consuming for y'all to study the entire population, i.eastward. all students in your college. Hence, you take out a sample of, say, 100 students and find out the boilerplate scores of those 100 students. This is the sample mean. Now, when you use this sample mean to infer about the population hateful, you won't be able to get the exact population means. In that location will exist some "margin of error".

You will now learn the answers to some of import questions: What is margin of error, what are the method of calculating margins of fault, how do you lot detect the critical value, and how to decide on t-score vs z-scores. Thereafter, you'll be given some margin of fault practice issues to make the concepts clearer.

What is Margin of Error?

The margin of error tin best exist described as the range of values on both sides (above and below) the sample statistic. For case, if the sample boilerplate scores of students are 80 and you make a statement that the average scores of students are 80 ± v, then here 5 is the margin of error.

Calculating Margins of Error

For calculating margins of mistake, you lot need to know the critical value and sample standard fault. This is because it's calculated using those two pieces of data.

The formula goes like this:

margin of error = critical value * sample standard error.

How do yous find the critical value, and how to calculate the sample standard error? Below, we'll discuss how to get these ii important values.

How exercise Yous discover the Critical Value?

For finding critical value, you need to know the distribution and the confidence level. For example, suppose you lot are looking at the sampling distribution of the ways. Hither are some guidelines.

If the population standard difference is known, apply z distribution.
If the population standard divergence is not known, utilise t distribution where degrees of liberty = northward-1 (n is the sample size). Notation that for other sampling distributions, degrees of freedom tin can be unlike and should be calculated differently using appropriate formula.
If the sample size is large, then apply z distribution (following the logic of Fundamental Limit Theorem).

It is important to know the distribution to decide what to use – t-scores vs z-scores.

Caution – when your sample size is large and it is not given that the distribution is normal, then by Central Limit Theorem, you can say that the distribution is normal and use z-score. All the same, when the sample size is small and it is not given that the distribution is normal, then y'all cannot conclude anything nearly the normality of the distribution and neither z-score nor t-score tin be used.

When finding the disquisitional value, confidence level will be given to you. If you are creating a ninety% conviction interval, then confidence level is 90%, for 95% conviction interval, the confidence level is 95%, and then on.

Here are the steps for finding disquisitional value:

Pace 1: Starting time, notice alpha (the level of significance). $\blastoff =one$ – Confidence level.

For 95% confidence level, $\blastoff =0.05$

For 99% confidence level, $\alpha =0.01$

Step ii: Find the critical probability $p*$ . Disquisitional probability will depend on whether we are creating a 1-sided confidence interval or a two-sided confidence interval.

For 2-sided confidence interval, $p*=1-\dfrac { \alpha }{ ii }$

For one-sided confidence interval, $p*=1-\blastoff$

Then you need to determine on using t-scores vs z-scores. Observe a z-score having a cumulative probability of $p*$ . For a t-statistic, notice a t-score having a cumulative probability of $p*$ and the calculated degrees of freedom. This will exist the critical value. To notice these disquisitional values, you should employ a calculator or respective statistical tables.

Sample Standard Error

Sample standard error can exist calculated using population standard deviation or sample standard difference (if population standard deviation is not known). For sampling distribution of means:

Allow sample standard departure be denoted past $s$ , population standard departure is denoted by $\sigma$ and sample size be denoted past $n$ .

$\text {Sample standard error}=\dfrac { \sigma }{ \sqrt { n } }$ , if $\sigma$ is known

$\text {Sample standard error}=\dfrac { south }{ \sqrt { n } }$ , if $\sigma$ is non known

Depending on the sampling distributions, the sample standard error can be different.

Having looked at everything that is required to create the margin of mistake, you can now straight calculate a margin of mistake using the formula we showed y'all earlier:

Margin of mistake = critical value * sample standard error.

Some Relationships

1. Conviction level and marginal of mistake

As the confidence level increases, the critical value increases and hence the margin of error increases. This is intuitive; the price paid for higher confidence level is that the margin of errors increases. If this was not so, and if higher confidence level meant lower margin of errors, nobody would cull a lower confidence level. In that location are e'er merchandise-offs!

2. Sample standard deviation and margin of error

Sample standard deviation talks about the variability in the sample. The more variability in the sample, the college the chances of fault, the greater the sample standard error and margin of error.

three. Sample size and margin of mistake

This was discussed in the Introduction section. It is intuitive that a greater sample size will be a closer representative of the population than a smaller sample size. Hence, the larger the sample size, the smaller the sample standard mistake and therefore the smaller the margin of error.

Margin of Error Practise Problems

Instance 1

25 students in their concluding year were selected at random from a loftier school for a survey. Among the survey participants, it was constitute that the boilerplate GPA (Grade Indicate Average) was 2.nine and the standard deviation of GPA was 0.5. What is the margin of error, assuming 95% confidence level? Requite correct interpretation.

Step one: Place the sample statistic.

Since you need to find the conviction interval for the population mean, the sample statistic is the sample mean which is the average GPA = two.9.

Step 2: Identify the distribution – t, z, etc. – and find the critical value based on whether you need a one-sided conviction interval or a two-sided confidence interval.

Since population standard deviation is not known and the sample size is small, use a t distribution.

$\text {Degrees of freedom}=northward-one=25-i=24$ .

$\alpha=1-\text {Conviction level}=1-0.95=0.05$

Permit the critical probability exist $p*$ .

For ii-sided confidence interval,

$p*=1-\dfrac { \blastoff }{ 2 } =i-\dfrac { 0.05 }{ 2 } =0.975$ .

The critical t value for cumulative probability of 0.975 and 24 degrees of freedom is 2.064.

Step iii: Find the sample standard error.

$\text{Sample standard error}=\dfrac { s }{ \sqrt { due north } } =\dfrac { 0.5 }{ \sqrt { 25 } } =0.1$

Step 4: Detect margin of error using the formula:

Margin of error = critical value * sample standard mistake

= 2.064 * 0.1 = 0.2064

Interpretation: For a 95% conviction level, the average GPA is going to be 0.2064 points in a higher place and below the sample average GPA of 2.9.

Example 2

400 students in Princeton University are randomly selected for a survey which is aimed at finding out the average time students spend in the library in a solar day. Among the survey participants, information technology was establish that the average time spent in the academy library was 45 minutes and the standard deviation was 10 minutes. Assuming 99% confidence level, find the margin of error and requite the correct interpretation of it.

Step 1: Place the sample statistic.

Since you demand to find the confidence interval for the population mean, the sample statistic is the sample mean which is the mean time spent in the university library = 45 minutes.

Stride 2: Identify the distribution – t, z, etc. and observe the critical value based on whether the need is a 1-sided confidence interval or a two-sided confidence interval.

The population standard departure is not known, but the sample size is large. Therefore, utilize a z (standard normal) distribution.

$\blastoff=ane-\text{Conviction level}=1-0.99=0.01$

Permit the critical probability be $p*$ .

For ii-sided confidence interval,

$p*=1-\dfrac { \alpha }{ 2 } =one-\dfrac { 0.01 }{ 2 } =0.995$ .

The critical z value for cumulative probability of 0.995 (every bit plant from the z tables) is ii.576.

Stride iii: Find the sample standard fault.

$\text{Sample standard error}=\dfrac { s }{ \sqrt { n } } =\dfrac { x }{ \sqrt { 400 } } =0.v$

Step 4: Find margin of error using the formula:

Margin of fault = disquisitional value * sample standard error

= 2.576 * 0.five = i.288

Interpretation: For a 99% confidence level, the mean time spent in the library is going to be ane.288 minutes above and below the sample mean time spent in the library of 45 minutes.

Instance 3

Consider a similar set up in Example 1 with slight changes. You randomly select X students in their final twelvemonth from a high school for a survey. Among the survey participants, information technology was found that the average GPA (Grade Point Boilerplate) was three.ane and the standard deviation of GPA was 0.vii. What should be the value of X (in other words, how many students you should select for the survey) if you lot want the margin of mistake to be at most 0.1? Assume 95% confidence level and normal distribution.

Step 1: Discover the critical value.

$\alpha=1-\text{Confidence level}=ane-0.95=0.05$

Let the critical probability exist $p*$ .

For two-sided confidence interval,

$p*=ane-\dfrac { \alpha }{ 2 } =1-\dfrac { 0.05 }{ 2 } =0.975$ .

The critical z value for cumulative probability of 0.975 is one.96.

Step 3: Find the sample standard fault in terms of $X$ .

$\text{Sample standard error}=\dfrac { s }{ \sqrt { 10 } }=\dfrac { 0.vii }{ \sqrt { X } }$

Pace four: Find $X$ using margin of error formula:

Margin of error = disquisitional value * sample standard error

$0.ane=1.96*\dfrac { 0.7 }{ \sqrt { 10 } }$

This gives $10=188.24$ .

Thus, a sample of 189 students should exist taken so that the margin of error is at most 0.1.

Determination

The margin of error is an extremely of import concept in statistics. This is because information technology is difficult to study the entire population and the sampling is not costless from sampling errors. The margin of error is used to create conviction intervals, and near of the fourth dimension the results are reported in the form of a confidence interval for a population parameter rather than just a single value. In this article, you lot made a beginning by learning answering questions like what is margin of fault, what is the method of calculating margins of errors, and how to interpret these calculations. You lot as well learned to decide whether to use t-scores vs z-scores and gained data about finding disquisitional values. Now you know how to utilise margin of error for constructing confidence intervals, which are widely used in statistics and econometrics.

Let'south put everything into practise. Endeavour this Statistics practice question:

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