How To Find Minimum Sample Size On Ti-84

How Large a Sample Do I Demand?

Summary: When you estimate a population parameter, you compute a conviction interval afterward taking sample information. Based on the confidence level that you lot preselect, and characteristics of your sample or population, compute a margin of fault.

Simply before you perform the study, how can you decide how big a sample y'all need and then that your confidence interval will take your desired margin of mistake or less?

The respond is that you lot take the formula for the margin of error, rearrange information technology algebraically to solve for the sample size, compute, and round upwards. This page shows the formulas for some common cases, with examples.

Case 0: One Population Mean, Known σ

If you know the standard divergence σ of the population, and you want to estimate the mean μ to within a given margin of mistake E in a ane−α confidence interval, here�s how to discover the required sample size n:

Due east = z _α/2 � σ/√ due north transforms to n = (z _α/two � σ/E)�

Case one: You lot want to estimate the boilerplate hourly output of a machine to within �1.five, with 90% confidence. Based on historical data, you have reason to believe that the standard deviation of the machine�s hourly output is six.two. How large a sample do you need?

Solution: Annotation starting time that this is not a realistic state of affairs. It�south pretty unlikely that you lot would know the standard divergence of a population merely non know the mean of that population. However, statistics texts always begin with this example because it�south the simplest way to demonstrate the principles. You go out Perfectland and enter Realityville in the other cases. With that said�

Comments	Computation
It�s good practice to showtime any problem by writing down what yous know and what yous demand, with symbols.	Given: E = 1.five, σ = vi.2, ane−α = 0.90. Wanted: sample size n
The formula wants z _α/2. How practise you compute it? Brainstorm by finding α/2.	ane−α = 0.xc ⇒ α = 0.x ⇒ α/two = 0.05 Since α/2 = 0.05, z _α/two =z _0.05
z _rtail is the critical z , or the z score that divides the normal bend leaving a right-manus tail with an surface area of rtail. Yous compute information technology on your TI-83/84/89 as `invNorm(i−` rtail `)`.	z _0.05 = invNorm(1−0.05) ≈ ane.6449
At present you accept all the pieces. Don�t utilize the rounded value of z _α/two, only use [ `2nd(-)` makes `ANS`] to continue full precision.	north = [ z _α/ii � σ � E ]� = (ANS�6.2�1.5)� = 46.2227... → 47

Answer: Given a population standard deviation of half-dozen.2 units per hr, if you take a sample size ≥47 the margin of mistake in a xc% conviction interval will be ≤1.5 units per hour.

Why do nosotros circular up? After computing 46.2227, why not report a sample size of 46? Well, the computation shows that a sample size of exactly 46.2227... would requite a margin of mistake of exactly 1.five. If you go slightly lower, to 46, the margin of error will exist slightly higher than 1.5. Since the sample size must exist a whole number, 46 or 47, and your margin of error must not exceed ane.5, you accept to choose the slightly higher number 47, which will give a margin of fault slightly less than 1.5.

Case 1: 1 Population Mean, Unknown σ

Note: Many bones statistics courses skip the material in this department and estimate sample sizes using a z distribution, and so the material in this section might exist an advanced extra for you lot. Check your class requirements.

This is the realistic case for estimating a population mean. Usually you don�t know the standard deviation of the population, so y'all have to use Student�southward t distribution instead of the normal (z) distribution. You approximate the standard deviation of the population from the standard divergence of a sample obtained in a prior study or a small-scale airplane pilot study. Here is the formula for sample size:

E = t _df,α/ii �s/√ due north transforms to northward = t _df,α/2 �s/E)�

There�s a certain element of Take hold of-22 in this formula for n. You don�t know n, so y'all don�t know the degrees of freedom df either and you tin�t compute the critical t for the formula. How do you go around this?

Utilise what NIST/SEMATECH calls an iterative method. Get-go compute the formula using z _α/2 instead of t _df,α/2. Then, when you have a preliminary sample size determined by (ab)using z in this way, recompute the formula using that sample size minus i for df. The two numbers should not be very different, since t is by and large not very unlike from z; but if they are, you tin apply the second number to compute t one time again.

Caution: It may happen that the formula gives a sample size less than 30. Only remember that your sample size must e'er exist at least thirty unless you lot have good reason to believe that the underlying population is normally distributed.

See also: Sample Sizes Required in NIST/SEMATECH e-Handbook of Statistical Methods: scroll downwards to �More often nosotros must compute the sample size with the population standard deviation being unknown�

Allow�due south illustrate the method using a modified form of the previous instance.

Example 2: You want to estimate the average hourly output of a auto to within �1.5, with 90% conviction. A small airplane pilot written report finds a sample standard deviation of the car�s hourly output is 6.2. How large a sample practice you need?

Solution: Apply z instead of t to brand a preliminary estimate, and so recompute with t.

Comments	Computation
Align your data.	Given: Eastward = i.5, s = 6.ii, 1−α = 0.90. Wanted: sample size n
The formula wants t _df,α/ii, simply nosotros approximate with z _α/2. Begin by finding α/2.	one−α = 0.xc ⇒ α = 0.10 ⇒ α/2 = 0.05 Since α/2 = 0.05, z _α/ii =z _0.05
z _0.05 is the disquisitional z-score that divides the normal distribution such that the surface area of the right-hand tail is 0.05, and therefore the area of the left-hand tail is 1−0.05.	z _0.05 = invNorm(ane−0.05) ≈ 1.6449
Now you lot have all the pieces you need for the preliminary sample size. Don�t use the rounded value of z _α/two, just utilize [ `2d(-)` makes `ANS`] to keep total precision.	due north = [ z _α/2 � south � Due east ]� = (ANS�six.ii�1.v)� = 46.2227... → 47
Your preliminary sample size is 47, and next you use that to compute t. df = n−1 = 46, so you lot need t _46,0.05. On the TI-84 you can utilize the `invT` office.	t _46,0.05 = 1.67866
At present recompute the formula using the t value. Remember, e'er circular sample size up.	due north = [ t _df,α/2 � s � Eastward ]� = (ANS�6.two�i.5)� = 48.142... → 49

Answer: Given a sample standard difference of six.2 units per hr, if you have a sample size ≥49 the margin of fault in a ninety% confidence interval will be ≤1.5 units per hour.

Remark: The sample size of 49 is a scrap larger than the Case 0 sample size of 47. This makes sense. When you don�t know the standard deviation of the population, you have to use the t distribution. Student�s t is more spread out than z, so the confidence intervals are a flake wider, then you have to utilize a larger sample to keep the confidence interval to the same width.

Example ii: One Population Proportion

For binomial data with true proportion p, the population standard departure is σ = √ p(1−p). Even though you don�t know p, the value p̂ (ane− p̂ ) from your sample � or your prior estimate, if you don�t take a sample � volition be shut to the true value p(one−p) in the population, because the production p(i−p) doesn�t vary much as p varies.

Replacing p with p̂ in a formula may seem similar adulterous, only n this instance it�due south not, because p(one−p) varies a lot less than p on its own. For instance, suppose the true population proportion is 45% but your estimate is 35%. The truthful p(i−p) is 0.45�0.55 = 0.2475, and your guess is 0.35�0.65 = 0.2275. The difference between 0.2475 and 0.2275 is a lot less than the departure between 0.45 and 0.35.

Therefore you tin can use a z role, and the formulas are the same every bit Case 0 with √ p̂ (i− p̂ ) substituted for σ:

E = z _α/2 � √ p̂ (one− p̂ )/n transforms to n = (z _α/2/E)� � p̂ (1− p̂ )

If y'all don�t have a sample or any credible estimate, use p̂ = (ane− p̂ ) = 0.5. This is the conservative process because the product p̂ (one− p̂ ) takes its highest value when p̂ = 0.5. The bourgeois procedure may give you a sample size larger than necessary, but y'all can be sure your sample won�t be too pocket-sized, forcing yous to throw out your survey and showtime over.

Caution: The sample must non exceed 10% of the population. Some other way to await at that is that 10 times sample size must be less than or equal to population size.

Example 3: What percent of the voters would vote for your candidate if the election were held today? You want 95% confidence in your answer, with a margin of mistake no more 3.five%. Last calendar month�s poll showed your candidate had 42% support. How many voters exercise you need to survey?

Comments	Computation
Marshal your data. Circumspection! iii.v% is 0.035, not 0.35.	Given: 1−α = 0.95, E = 0.035, p̂ = 0.42 Wanted: sample size northward
To discover z _α/two, offset find α/2.	1−α = 0.95 ⇒ α = 0.05 ⇒ α/2 = 0.025
z _α/2 = z _0.025, the critical z for a right-hand tail area of 0.025. That�southward invNorm(ane−.025).
Carve up by E and square. You lot�re going to chain calculations so that yous don�t have to re-enter any of your intermediate numbers. Press [`/`], and notice how the figurer responds `Ans` to allow you know it�s using the previous answer. Enter .035 for E and press [`ENTER`]; that gives yous the result of the fraction. Press [`x�`] [`ENTER`] to foursquare it.
The last link in the chain is multiplying past p̂ and and so by (1− p̂ ). Your result is 764. Recollect, always round sample size upwards, regardless of the decimal role.

Respond: To notice a 95% CI with a margin of error no more than �three.5 percentage points, where the true population proportion is around 42%, you lot must survey at least 764 people.

10�764 = 7640; presumably the electorate is larger than that.

Example 4: Suppose you�re planning your kickoff poll, and you have no idea of your candidate�s level of back up. How large a sample would yous need to exist sure of a margin of error no more than than 3.5% in a 95% CI?

Solution: Compute z _α/2 = ane.9600 as in the previous case. But this time apply p̂ = 0.5 since y'all have no estimate for p.

northward = ( z _α/2 � E )� p̂ (1− p̂ ) = (invNorm(ane-.025)/.035)� � 0.42 � (i−0.42) = 783.971... → 784

Answer: To find an 95% CI with a margin of error no more than �3.v percentage points, where you have no idea of the truthful population proportion, y'all must survey at least 784 people.

Example 5: Difference of Two Population Proportions

When you lot�re comparing 2 population proportions, it�s perfectly legitimate to have different-sized samples. The formula for margin of mistake, below left, is just an extension of the formula for one population proportion.

Simply when y'all�re planning sample size, you can�t solve one equation for two variables n ₁ and n ₂. (If y'all had a reason to choose some particular value for i of them, you could solve for the other one.) Yous can solve for sample size if yous determine to utilise the same size for both samples.

$E equals [z of alpha over 2] times square root of fraction p1 times 1 minus p1 over n1 plus fraction p2 times 1 minus p2 over n2$ transforms to
north = due north _ane = due north ₂ = [ p̂ _one(i− p̂ ₁) + p̂ _ii(1− p̂ _ii)] � (z _α/2/Eastward)�

For the reasons given higher up, if y'all take any prior estimates for the population proportions p ₁ and p ₂ you should use them; otherwise use 0.v.

Example 5: You�d like to know how your candidate�s support differs between men and women. Y'all know that overall back up is 42%. How many of each sex must yous survey to answer the question with 90% confidence and a margin of error no more than 3%?

Comments	Computation
Marshal your data. (Caution! 3% is 0.03 not 0.three.)	Given: 1−α = 0.ninety, E = 0.03 Wanted: sample size northward=n _i=n _ii
Practise you have an estimate of p _ane and p ₂? Yes, since the overall support is 42% yous expect that men�southward and women�southward support is not as well different from that. (Yous do expect p _i and p _two are somewhat different, or you wouldn�t be doing the survey. But remember from one population proportion that p(1−p) doesn�t vary much when p varies.)	Prior: p̂ ₁ = 0.42 and one− p̂ ₁ = 0.58 p̂ _two = 0.42 and ane− p̂ ₂ = 0.58
Compute z _α/2 in the usual way.	i−α = 0.90 ⇒ α = 0.10 ⇒ α/two = 0.05 z _0.05 = invNorm(1−0.05) ≈ 1.6449
Finish, using the unrounded value of z. Always remember to circular sample sizes upwardly.	[ p̂ ₁(1− p̂ ₁) + p̂ ₂(1− p̂ ₂)] [z _α/ii�E]� = (0.42�0.58+0.42�0.58)�(ANS�0.03)� = 1464.60... → 1465

Answer: To find a ninety% CI for the difference in your candidate�s support between men and women, with margin of error no more than than 3%, you lot must survey at to the lowest degree 1465 men and at least 1465 women.

Remark: You might wonder why the samples must be so large. After all, to estimate one population proportion to �3% in a 90% CI, with prior gauge p̂ = 42%, a sample of 752 is enough. (Check information technology!) Why do you need over 2900 people in two groups for the aforementioned margin of error?

The answer is that it�s not the same margin of error. If you surveyed 752 men and 752 women you�d accept confidence intervals of �three% for each, but that�s an overall margin of fault of �6% � think that the true proportion might be well-nigh the bottom of 1 group�s interval and near the top of the other group�s, or vice versa. (It�south not quite that simple, but that�s the bones thought.) To bring downwards that margin of error, you take to increment the sample size.

Example 6: Permit�due south modify the previous example. Suppose yous have reason to believe your candidate appeals more strongly to women, with a gap of about 10%? That means you lot estimate men�s support at 37% and women�southward support at 47%. p̂ ₁ = 0.37, i− p̂ _ane = 0.63, p̂ ₂ = 0.47, 1− p̂ ₂ = 0.52. Your required sample size becomes

z _0.05 = invNorm(1−0.05) ≈ 1.6449

[ p̂ ₁(1− p̂ _one) + p̂ ₂(1− p̂ ₂)] [z _α/two�E]� = (0.37�0.63+0.47�0.53)�(ANS�0.03)� = 1449.57... → 1450

Answer: For a 90% CI with margin of error ≤3%, when you recollect one population�southward proportion is 37% and the other�s is 47%, you demand a sample of at to the lowest degree 1450 from each group.

Instance 6: Goodness of Fit

For χ� tests, the requirements are that all of the expected counts must be ≥5. The expected count for each category is sample sizedue north times the model proportion in that category, so to find the the necessary sample size you divide that minimum expected value of 5 past the smallest proportion or percentage in the model:

n = five / (the smallest proportion in your model)

If your model is expressed in ratios, such as 9:3:3:1, the smallest proportion is the smallest number in the model divided past the total in the model, in this case 1/(9+3+3+1) or 1/16. So the required sample size is 5 / (1/16) = 80.

Example 7: Yous expect that customers will choose coffee, tea, bottled h2o, and Snapple in the proportions of 65%, 15%, 15%, five%. How large a random sample must you accept to test this model?

Solution: Take the least probable category and divide 5 by that percentage:

n = 5 / v% = 5 / 0.05 = 100.

Respond: Y'all need a random sample of at least 100 to test this model. (Every bit always, the minimum sample will requite a significant event merely if the null hypothesis is extremely incorrect. If the model is but moderately wrong, a larger sample volition probably be needed to reveal that.)

Instance 8: For a kids� play area, you ordered five cartons of blue plastic assurance, two of green, and three each of red and yellow. But your banana dumped them all together after taking delivery. How many balls must you select randomly to see if the proportions are right?

Solution: You must carve up v by the proportion in the smallest category in the model, which is light-green:

smallest proportion = 2 / (5+2+iii+three) = 2/13

v / (2/thirteen) = 32.5 → 33

Answer: To see the requirements, you need a random sample of at to the lowest degree 33 balls.

Example 9: Y'all believe that patently M&Ms are distributed in the proportions 24% blue, xiii% brown, 16% green, eighteen% orange, fifteen% red, 14% yellow. How large a sample exercise y'all need to examination this model?

Solution: The smallest proportion in the model is 13%, and so compute 5/13% = 5/0.xiii = 38.46... → 39. (Remember, sample sizes e'er round up.)

Answer: The sample must comprise at least 39 M&Ms.

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