One problem with the split-half method is that the reliability estimate obtained using any random split of the items is likely to differ from that obtained using another. One solution to this problem is to compute the Spearman-Brown corrected split-half reliability coefficient for every one of the possible split-halves and then find the mean of those coefficients. This is the motivation for Cronbach’s alpha.

Cronbach’s alpha is superior to Kuder and Richardson Formula 20 since it can be used with continuous and non-dichotomous data. In particular, it can be used for testing with partial credit and for questionnaires using a Likert scale.

**Definition 1**: Given variable *x*_{1}*, …, x _{k}* and

*x*

_{0}= and

**Cronbach’s alpha**is defined to be

**Property 1**: Let *x _{j} = t_{j} + e_{j}* where each

*e*is independent of

_{j}*t*and all the

_{j}*e*are independent of each other. Also let

_{j}*x*

_{0}= and

*t*

_{0}= . Then the reliability of

*x*

_{0}≥

*α*where

*α*is Cronbach’s alpha.

Here we view the *x _{j}* as the measured values, the

*t*as the true values and the

_{j}*e*as the measurement error values. Click here for a proof of Property 1.

_{j}**Observation**: Cronbach’s alpha provides a useful lower bound on reliability (as seen in Property 1). Cronbach’s alpha will generally increase when the correlations between the items increase. For this reason the coefficient measures the internal consistency of the test. Its maximum value is 1, and usually its minimum is 0, although it can be negative (see below).

A commonly-accepted rule of thumb is that an alpha of 0.7 (some say 0.6) indicates acceptable reliability and 0.8 or higher indicates good reliability. Very high reliability (0.95 or higher) is not necessarily desirable, as this indicates that the items may be entirely redundant. These are only guidelines and the actual value of Cronbach’s alpha will depend on many things. E.g. as the number of items increases, Cronbach’s alpha tends to increase too even without any increase in internal consistency.

The goal in designing a reliable instrument is for scores on similar items to be related (internally consistent), but for each to contribute some unique information as well.

**Observation**: There are an number reasons why Cronbach’s alpha could be low or even negative even for a perfectly valid test. Two such reasons are reverse coding and multiple factors.

**Reverse coding**: Suppose you use a Likert scale of 1 to 7 with 1 meaning strongly disagree and 7 meaning strongly agree. Suppose two of your questions are: Q1: “I like pizza” and Q20: “I dislike pizza”. These questions ask the same thing, but with reverse wording. In order to apply Cronbach’s alpha properly you need to reverse the scoring of any negatively phrased question, Q20 in our example. Thus if a response to Q20 is say 2, it needs to be scored as 6 instead of 2 (i.e. 8 minus the recorded score).

**Multiple factors**: Cronbach’s alpha is useful where all the questions are testing more or less the same thing, called a “factor”. If there are multiple factors then you need to determine which questions are testing which factors. If say there are 3 factors (e.g. *happiness with your job, happiness with your marriage* and *happiness with yourself*), then you need to split the questionnaire/test into three tests, one containing the questions testing factor 1, one with the questions testing factor 2 and the third with questions testing factor 3. You then calculate Cronbach’s alpha for each of the three tests. The process of determining these “hidden” factors and splitting the test by factor is called Factor Analysis (see Factor Analysis).

**Example 1**: Calculate Cronbach’s alpha for the data in Example 1 of Kuder and Richardson Formula 20 (repeated in Figure 1 below).

The worksheet in Figure 1 is very similar to the worksheet in Figure 1 of Kuder and Richardson Formula 20. Row 17 contains the variance for each of the questions. E.g. the variance for question 1 (cell B17) is calculated by the formula =VARP(B4:B15). Other key formulas used to calculate Cronbach’s alpha in Figure 1 are described in Figure 2.

**Figure 2 – ****Key formulas for the worksheet in Figure 1**

Since the questions only have two answers, Cronbach’s alpha .73082 We see that this is the same as the We see that this is the same as the KR20 reliability calculated for Example 1 of Kuder and Richardson Formula 20.

**Observation**: If the variances of the *x _{j}* vary widely, the

*x*can be standardized to obtain a standard deviation of 1 prior to calculating Cronbach’s alpha.

_{j}**Observation**: To determine how each question on a test impacts the reliability, Cronbach’s alpha can be calculated after deleting the *i*th variable, for each *i ≤ k.* Thus for a test with *k* questions, each with score *x _{j}*, Cronbach’s alpha is calculated for for all

*i*where = .

If the reliability coefficient increases after an item is deleted, you can assume that the item is not highly correlated with the other items. Conversely, if the reliability coefficient decreases, you can assume that the item is highly correlated with the other items.

**Example 2**: Calculate Cronbach’s alpha for the survey in Example 1, where any one question is removed.

The necessary calculations are displayed in Figure 3.

Each of the columns B through L represents the test with one question removed. Column B corresponds to question #1, column C corresponds to question #2, etc. Figure 4 displays the formulas corresponding to question #1 (i.e. column B); the formulas for the other questions are similar. Some of the references are to cells shown in Figure 2.

**Figure 4 – ****Key formulas for worksheet in Figure 3**

As can be seen from Figure 3, the omission of any single question doesn’t change the Cronbach’s alpha very much. Removal of Q8 affects the result the most.

**Observation**: Another way to calculate Cronbach’s alpha is to use the **Two Factor ANOVA without Replication** data analysis tool on the raw data and note that:

**Example 3**: Calculate the Cronbach’s alpha for Example 1 using ANOVA.

We begin by running Excel’s **Anova: Two Factor without Replication** data analysis tool using the data in range B4:L15 of the worksheet shown in Figure 1.

As you can see from Figure 5, Cronbach’s alpha is .73802, the same value calculated in Figure 1.

**Observation**: Alternatively, we could use the Real Statistics **Two Factor ANOVA** data analysis tool, setting the **Number of Rows per Sample** to 1. We can also obtain the same result using the following supplemental function.

**Real Statistics Function**: The following function is provided in the Real Statistics Resource Pack:

**CRONALPHA**(R1) = Cronbach’s alpha for the data in range R1

As noted above, for the data in Figure 1, CRONALPHA(B4:L15) = .738019.

**Example 4**: Calculate Cronbach’s alpha for a 10 question questionnaire with Likert scores between 1 and 7 based on the 15 person sample shown in Figure 6.

As you can see from Figure 6, Cronbach’s alpha is 0.59172, a little below the generally acceptable range. We get the same answer by using the supplemental formula in the Real Statistics Resource Pack, namely CRONALPHA(B4:K18) = 0.59172.

the test contains 18 questions in which 12 mcq’s,5 two marks questions and 1 three marks questions ,how can i calculate alpha coefficient

Hi Jeena,

I believe that for each question you simply score 1 for a correct answer and 0 for an incorrect answer, whether the multiple choice question has 2,3 or 4 choices. Then follow the procedure described in Example 1 or Example 3 on webpage http://www.real-statistics.com/reliability/cronbachs-alpha/. You can also use the supplemental formula CRONALPHA provided in the Real Statistics Resource Pack.

Charles

how can i calculate five point scale survey in excel and get the cronbach alpha??? its quite confusing.

Mary Ann,

Enter your data as in Example 4 on http://www.real-statistics.com/reliability/cronbachs-alpha/. The example shows what to do for a seven point scale, but a five point scale works exactly the same. Then carry out the calculations as in Figure 6 on the same webpage or simply use CRONALPHA(R1) where R1 is the the range containing your data (without headings). CRONALPHA is a supplemental function which is contained in the Real Statistics Resource Pack.

Charles

Thank you, Sir. It´s a great example. it contains all that I have searched. ¿Aren´t you teacher?

Hi. I got negative cronbach alpha which made it unacceptable but the questions are just right and fit to my study. What should I do? Thank you for your response. God bless.

Hi Krsna,

There are an number reasons for getting a low (or even negative) value for cronbach’s alpha for a perfectly valid test. Two reasons are reverse coding and multiple factors.

Reverse coding: Suppose you use a Likert scale of 1 to 7 with 1 meaning strongly disagree and 7 meaning strongly agree. Suppose two of your questions are: Q1: I like pizza and Q20: I dislike pizza. These questions ask the same thing, but with reverse scoring. In order to apply Cronbach’s alpha properly you need to reverse the scoring of the negatively phrased question, Q20. Thus if a response to Q20 is say 2, it needs to be scored as 6 instead of 2 (i.e. 8 minus the score).

Multiple factors: Cronbach’s alpha is useful where all the questions are testing more or less the same thing, called a “factor”. If there are multiple factors then you need to determine which questions are testing which factors. If say there are 3 factors, then you need to split the questionnaire/test into three tests, one containing the questions testing factor 1, one with the questions testing factor 2 and the third with questions testing factor 3. You then calculate cronbach’s alpha for each of the three tests. The process of determining these “hidden” factors and splitting the test by factor is called Factor Analysis. See the webpage http://www.real-statistics.com/multivariate-statistics/factor-analysis/ for more details about how to do this in Excel using Real Statistics.

Charles

this method was very helpful for an average student like me,thank you very much…

Hi,

Thank you for the great explanation.

My work includes using a 5 point scale by 2 separate groups of raters (trained Vs untrained in that field). Each group will rate 20 different cases.

My questions are:

1- Do I need to use Cronbach’s alpha for each case separately (20 cases x 2 groups) (i.e. 40 times)?

2- How can I assess the consistency within each group for all the cases collectively?

3- How can I know if there is a significant difference between the 2 groups for all the cases collectively?

Apologies for the long questions, but I really appreciate your help.

Best regards,

Dalia

Hi Dalla,

From what you have described I have the following suggestions:

1. to assess the agreement/disagreement between the two raters you probably want to use something like Cohen’s kappa

2. to assess the consistency within each of the two groups you can use Cronbach’s alpha

3. to assess whether there is a significant difference between the two groups you can use either a t test or Mann-Whitney

All of these tests are described on the Real Statistics website and the Real Statistics Resource Pack can be used to carry out each of these tests.

Charles

Hi,

I have approximately 200 respondense, could you please let me know how to ho about getting the results of the Cronbach test. Do I have to enter a figure for each respondent to a question?

Thanks,

MC.

Mukesh,

If you have say 200 people taking the test and each test has 20 questions, then you would create a range similar to that in Figure 1 with 200 rows and 20 columns. You need to enter a figure to each question.

Charles

Great site. Thank you.

I am looking for simple consistencies within a singular column of measured data, to see how volatile it is. I tried to use your CRONALPHA function to simply analyze one column of data of designated cells, and am getting the “#VALUE” response.

Am I missing something? Thanks much.

Jonathan

Jonathan,

Cronbach’s alpha is not designed to do what you want. There is no internal consistency to measure when only have one response per subject. To use CRONALPHA you need at least two columns.

Charles

Thank you Charles.

Perhaps I am asking the wrong question. Let’s say I have 120 responses from one subject over time, and what I am trying to do is to look at the consistency and repeatability of the answers from that subject. In other words, I want to see if the data, once I reach a certain point, has a stability to it. While I could do an even-odd correlation from the set, 1st-half / 2nd-half, or something like that, that introduces the limitations you discuss.

Is there something that considers the stability of a great number of data from one subject alone, to look at stability purposes? Thanks very much for a great site.

Jonathan,

I am afraid that I don’t have much help to offer. I checked the Internet and found a number of sites addressing

statistical analysis of single subject dataandsingle subject consistency statistics, including the following:http://www.physther.net/content/74/8/768.full.pdf

http://www.u.arizona.edu/~dusana/psych290Bpresession06/notes/Ch%2014%20Single-Subejct%20Research%20Designs.ppt

Unfortunately these sites seem to be addressing something different from what I think you are looking for.

Charles

If I have a 2 group study and want to conduct a cronbach alpha test, do I have to separate it into 2 groups?

Mabel,

It really depends on what you want to study. E.g. if the groups are just a random split of the sample you could use the split-half method instead of Cronbach’s alpha to measure reliability.

Charles

Thank you Charles. Your response is highly appreciated. My apology for this late reply. God bless.

Do you have a reference for this”A commonly-accepted rule of thumb is that an alpha of 0.6-0.7 indicates acceptable reliability”? Thank you very much!

Vien,

I believe that I got this from the following reference (this is also referenced in Wikipedia)

Ref 1: Kline, P. (2000). The handbook of psychological testing (2nd ed.). London: Routledge, page 13

I have generally seen that .7 is viewed as the minimum acceptable level. Here is such a reference.

Ref 2: George, D., & Mallery, P. (2003). SPSS for Windows step by step: A simple guide and

reference. 11.0 update (4th ed.). Boston: Allyn & Bacon. p. 231

The rules of thumb there are:

> .9 – Excellent, _ > .8 – Good, _ > .7 – Acceptable, _ > .6 – Questionable, _ > .5 – Poor, and < .5 – Unacceptable

Here is yet another example where .6 is used as the minimal acceptable level.

Ref 3: "Cronbach’s alpha (Cronbach, 1951) which quantifies the degree of internal consistency (reliability) of a set of items, was calculated for each subscale, as well as the overall scale. In general, a Cronbach’s alpha of at least .7 is the criterion used to establish an acceptable level of reliability. However, the recommended minimum Cronbach’s alpha for exploratory studies is .6" (Nunnally, J.C. (1978). Psychometric Theory (2nd ed.). New York: McGraw Hill; Robinson, J. P., Shaver, P. R., & Wrightsman, L. S. (Eds.). (1991). Measures of personality an social psychological attitudes. San Diego: Academic Press).

I’m not a student of mathematics. I am leading a research in applied linguistics and I need to calculate a result using Cronbach alpha. I asked my students about the extent to which they became autonomous after introducing language learning strategies on a five-graded Likert scale with scores from 1 to 5. However, I didn’t know how to calculate. On the vertical column, I have to mention the respondents (students) and on the horizontal line, I have only one question (so one item??). Could I calculate alpha with only one column?

I am seeking guidance and I would be very grateful if you could help me.

Thank you.

Hadia,

You can’t measure consistency between the items (which is what Cronbach’s alpha does) since you only have one item. You need more than one item to use Cronbach’s alpha. My question to you is why do you want to use Cronbach’s alpha? What are you trying to demonstrate?

Charles

Hi,

I’m using a sample of 40 and I have 2 sets of questions, one with 9 and the other with 10 questions. I’ve calculated the alpha for both using the formula as you explained and then using the Real Statistics toolpack. I’m getting different alphas when I use the two methods with the toolpack function giving an alpha higher by around 0.08. Why is that?

Karen,

Please email me the Excel worksheet(s) so that I can figure out happened.

Charles

how did you get the var in example 4. i just don’t get it how it can be computed. is there a formula needed?

Krisliz,

Cell B24 contains the formula =VARP(L4:L18).

Charles

Hello,

I am using an instrument that has been previously tested for reliability with a Cronbach Alpha of .91 and .93 for overall instrument (there are two variables being tested). The instrument has four subscales and each subscale has a Cronbach Alpha number of >.61 (three subscales are in the .80 range). I used the same instrument and tested Cronbach Alpha and received an overall number of .87 and .86. All of my subscales are substantially lower than the original testing by the instrument author. How do I explain that? We used a similar population. Her study was larger with 64 participants and mine had 40.

Thank you.

Colleen

Colleen,

A few reasons why your alpha values may differ are: population not really the same as that for the original instrument, randomness (you say that your values were substantially lower, but it may turn out that the difference is not statistically significant), differences in the way the test was administrated(e.g. a noisy environment, people under stress, etc.).

Charles

Hello,

I am having trouble with measures that has items with reverse coding. When I calculate the data with reverse coding, the Cronbach’s Alpha is very low (.2), however when I calculate the Cronbach’s Alpha without using the reverse coding, it is very high (.9). I am wondering if you know why that happens, and what I should do?

Dianna,

It is hard for me to tell without seeing the data, but I can think of the following possibilities:

- the questions that you have identified as reverse coded are not really reverse coded

- there is an error in the coding or calculation

- reliability is low (not sure why alpha would be high if reverse coding is not done)

Charles

Regarding example number 3, I have performed Anova: Two Factor without Replication in Excel, but could not see Alpha value as it has been shown in Figure 5 above.

Zargoon,

I just rechecked and I believe it is correct. Please make sure you analyzed the data from Example 1 (0′s and 1′s only) and not from Example 2.

Charles

Hi there – hopefully you can answer me

I would like to run a Cronbachs alfa test (or similar if you have an alteriour suggestion) to include it in my reliability part. (using SPSS)

I have a mixture of variables – Nominal, Ordinal and scale.

Which have a whole range of diferent ranges in values. Some variables have 0-1, others have 1-5, while others go from 0-14000. None are similiarly kategorized. Like, sex is a 0-1. Work hours range pr week from 0-30. A satisfactory question ranges from 1-5, rent payed ranges from 500- 13000… etc..

I’ve kategorized rent, and when I lump them all togheter in SPSS – Cronbachs alfa I get a Cronbachs Alfa of around ,500.

BUT – I have a feeling this is quite pointless? Because of the state of the values – is this so, is Cronbachs Alfa usless – how would I explain that I can’t use Cronbachs Alfa? My corriculum states that it is restricted to indexes – But I can’t quite understand what this means by “indexes”. Certainly we have an index of questions that all aim to answer an underlying question – but it’s not an index where all numbers are grouped simillarilly – so i am unsure what it means… Can’t I use it ?

Colene,

I have read a lot of conflicting information regarding the subject you are raising and so I don’t have a precise answer for you. My understanding is that Cronbach’s alpha is most relevant when the test is evaluating a single “factor”. You can certainly calculate Cronbach’s alpha even if the questionnaire contains a mix of multiple choice, true-false and other types of questions. What I would be especially cautious about is when the test is performing different types of evaluations (e.g. Likert scale to assess your satisfaction with a product plus multiple choice to assess your ability to use the product). You definitely need to calculate separate Cronbach alpha for each concept/factor that you are testing.

I’m nor sure what your curriculum means by “indexes”. Cronbach’s alpha itself can be used as an index, but it doesn’t sound like this is what is meant.

The following are a couple of articles on the web that may give you further information (although they may confuse things even more). I suggest that you speak to your professor to get further insights from him/her. Please share with the rest of us any insights you glean.

http://www.ctbassessments.com/assessment_insights/february_2011/research_insights.html

http://psychweb.psy.umt.edu/denis/datadecision/front/cortina_alpha.pdf

Charles

Thank you very much for this website!! Absolutely excellent and very useful…!

hi. i gor high Cronbach for my 4 variables but the correlations are zero. Is it somehing wrong wih my calculations or the data? Need help. Tq

Dee,

Can you send me a file with the example so that I can take a look at it?

Charles

Hye, i would like to run cronbach alpha on each item as my on likert scale..as you show in example 4 that run crunbach alpha on all the items..

How can i do it on each item?

Possible?

Seems it does not clear with me here..

Thanks

I’m not sure what you mean by “run cronbach on each item”. You run cronbach on all the items. You can also run cronbach on all the items except one (as in Example 2). In any case, how you calculate cronbach is described on the referenced webpage and you can get more information by looking at the example worksheet which you can download for free at webpage http://www.real-statistics.com/free-download/real-statistics-examples-workbook/.

Charles