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Data Scientist, Analytics was asked...6 March 2015

Write a SQL query to compute a frequency table of a certain attribute involving two joins. What if you want to GROUP or ORDER BY some attribute? What changes would you need to make? How would you account for NULLs?

24 Answers

If you group by parent_id, you'll be leaving out all posts with zero comments.

@ RLeung shouldn't you use left join? You are effectively losing all posts with zero comment. Less

Here is the solution. You need a left self join that accounts for posts with zero comments. Select children , count(submission_id) from ( Select a.submission_id, count(b.submission_id) as children from Submissions a Left Join submissions b on On a.submission_id=b.parent_id Where a.parent_id is null Group by a.submission_id ) a Group by children Less

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There are two mobile restroom stalls at a construction site where I work. There are also three situations that have an equal chance of occurrence: a. none of them is occupied b. only one of them is occupied c. both are occupied 1. If I were to pick one at random, what is the probability that it is occupied? 2. If it turns out that that first one I go to is occupied and I decide to try the other one, what is the probability that the second one is also occupied?

13 Answers

the answer to the first question is - 1/3 + 1/3*1/2 = 1/2 the answer to the second question require the formula of conditional probability. Let's say: P(A) - probability that second stall is occupied P(B) - probability that the first stall is occupied P(A\B) = P(AandB) / P(B) P(B) = 1/2 (first question) P(AandB) = 1/3 P(A\B) = (1/3) / (1/2) = 2/3 Less

Above answer is wrong, the answer to the first and second question are both 1/2.

there are 2 doors (A and B) and related 4 options with given probabilities option 1: A is full ; B is full ; prob 1/3 option 2: A is full; B is empty; prob "m" option 3: A is empty; B is full; prob "n" option 4: A is empty, B is empty, prob 1/3 we know m +n = 1/3 First question I have p1 probability to chose A and 1-p1 probability to chose B * I chose A and A is full: prob = 1/3+m * I chose B and B is full: prob = 1/3+n -> all together p1 *(1/3+m) + (1-p1)(1/3+n) -> assuming p1 =1/2 (probability of choosing door A and B equal) this gives 1/2 by using m+n = 1/3 Second question I chose door A first; prob of it being full is (1/3+a); given it is full we have 2 options "1" and "2". * Given A full being B is full (1/3)(1/3+a) * Given B full being A is full (1/3)(1/3+b) = (1/3)(1/3+(1/3-a)) then the total prob will be p1 * (1/3)(1/3+a) + (1-p1)(1/3)(1/3+(1/3-a)) if p1 =1/2 doors are not different in prob to be full if a=b = 1/6 choosing doors by the person is equally probable then this equation give 2/3 say door a is closer to the concert hall and b never gets full if A is empty and you always test B first; than p1 = 0; a = 1/3; b = 0 than If you find B full don't test A because the equation gives 1 Less

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Lets say the population on Facebook clicks ads with a click-through-rate of P. We select a sample of size N and examine the sample's conversion rate, denoted by hat{P}, what is the minimum sample size N such that Probability( ABS(hat{P} - P) < DELTA ) = 95%. In other words (this is my translation), find the minimum sample size N such that our sample estimate hat{P} is within DELTA of the true click through rate P, with 95% confidence.

6 Answers

Interpret the question this way: we want to choose an N such that P_hat is an element of [P - delta, P + delta] with probability 95%. First, note that since P_hat is the sum of N Bernoulli trials with some common parameter (by assumption) that we are trying to estimate, we can safely assume P_hat to be normally distributed with mean equal to the true mean (P) and variance equal to (P)(1 - P) / N. Now, we when does a normally distributed random variable fall within delta of it's mean with 95% probability? The answer depends on how big delta is. Since P_hat is normally distributed, we know from our statistics classes that 95% of the time it will fall within 2 standard deviations of its mean. So in other words, we want [P - delta, P + delta] = [P - 2*SE(P_hat), P + 2*SE(P_hat)]. That is, we want delta = SE(P_hat). So what is the SE ("standard error") of P_hat? Well that's just the square root of its (sample) variance, or Sqrt(P_hat * (1 - P_hat) / N). But wait! We haven't run the experiment yet! How can we know what P_hat is? We can either (a) make an educated guess, or (b) take the "worst" possible case and use that to upper bound N. Let's go with option (b): P_hat * (1 - P_hat) is maximized when P_hat is .5, so the product is 0.25. To put it all together: delta = 2 * Sqrt(0.25) / Sqrt(N) = 2 * .5 / Sqrt(N) => N = (1 / delta) ^ 2. So when N is greater than (1 / delta)^2, we can rest assured that P_hat will fall within the acceptable range 95% of the time. Less

Use Chebyshev's inequality

Rate has Poisson distribution, not Bernoulli. The mean equals the variance, SE = sqrt(P/N). Less

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Given two binary strings, write a function that adds them. You are not allowed to use any built in string to int conversions or parsing tools. E.g. Given "100" and "111" you should return "1011". What is the time and space complexity of your algorithm?

6 Answers

In Python: def normalize_length(str1, str2): len1 = len(str1) len2 = len(str2) if (len1 = 0): if (input2[i] == "1") and (input1[i] == "1"): if(carry): result = "1" + result carry = 1 else: carry = 1 result = "0" + result i -= 1 if (input2[i] == "1") and (input1[i] == "0"): if (carry): result = "0" + result else: result = "1" + result i -= 1 if (input2[i] == "0") and (input1[i] == "1"): if (carry): result = "0" + result else: result = "1" + result i -=1 if (input2[i] == "0") and (input1[i] == "0"): if (carry): result = "1" + result carry = 0 else: result = "0" + result i -=1 if(carry): result = "1" + result carry = 0 return(result) str1 = "111" str2 = "1011" print(normalize_length(str1, str2)) print(add_binary(str1, str2)) Obviously there are better ways to do this, but hey: my solution is O(N). Less

Ignore the answer above - didn't realize that Glassdoor would cut off parts of my answer for being too long. Assuming you already wrote the normalizing code to make the input lengths the same by adding zeros: def add_binary(input1, input2): normalized = normalize_length(input1, input2) input1 = normalized[0] input2 = normalized[1] length = len(input1) result = "" carry = 0 i = length-1 while(i >= 0): if (input2[i] == "1") and (input1[i] == "1"): if(carry): result = "1" + result carry = 1 else: carry = 1 result = "0" + result i -= 1 if (input2[i] == "1") and (input1[i] == "0"): if (carry): result = "0" + result else: result = "1" + result i -= 1 if (input2[i] == "0") and (input1[i] == "1"): if (carry): result = "0" + result else: result = "1" + result i -=1 if (input2[i] == "0") and (input1[i] == "0"): if (carry): result = "1" + result carry = 0 else: result = "0" + result i -=1 if(carry): result = "1" + result carry = 0 return(result) Less

def calc_bin_sum(bin1, bin2): ## bin1 conversion to a number based in 10 b1 = 0 for i in range(len(bin1)): b1 = b1 + int(bin1[i]) * (2**i) ## bin2 conversion to a number based in 10 b2 = 0 for j in range(len(bin2)): b2 = b2 + int(bin2[j]) * (2**i) ## Add two numbers corr_based_10 = b1 + b2 ## Change it back to binary def trans(x): binary = [] while x: binary.append(x % 2) x >>= 1 return binary return ''.join(map(str, trans(corr_based_10))) Less

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We have a table called ad_accounts(account_id, date, status). Status can be active/closed/fraud. A) what percent of active accounts are fraud? B) How many accounts became fraud today for the first time? C) What would be the financial impact of letting fraud accounts become active (how would you approach this question)?

6 Answers

A) what percent of active accounts are fraud? Select sum(Case when status = ‘fraud’ then 1 else 0 end)/count(*) as Fraud_percentage from ad_accounts where status ‘closed’; B) How many accounts became fraud today for the first time? select count(*) from ( select account_id, min(date) as First_fraud from ad_accounts where status = 'fraud' group by account_id having First_fraud = current_date() ); Less

Yep, should be A) what percent of active accounts are fraud? SELECT COUNT(DISTINCT t2.account_id)/COUNT( DISTINCT t1.account_id) AS perc_fraud FROM ad_accounts AS t1 LEFT JOIN ad_accounts AS t2 ON t1.account_id = t2.account_id AND t2.status = 'fraud' AND > WHERE t1.status = 'active' Less

For question B, if I assume i have today's data ans yesterday's data in the table, would this work? Select Count (distinct a.Account_id) From ad_accounts A Inner join ad_accounts b On a.account_id=b.account_id Where and (‘day’, -1, current_date) And a.status=’fraud’ And b.status!=’fraud’ Less

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Hyundai Capital America

what is logistic regression? How to perform variable selection

5 Answers

See answers for Capital One statistician questions

A Measurement of Variables

Logistic regression is a predictive analysis. To explain the relationship between the one dependent variable with another independent variable. Less

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How will you develop a method by Chromatography for a high matrix ed samples?

4 Answers

trial and error

Depends on the type of matrix . If proteins present, then electrophoretic method will be good but it's time consuming. If other species, then column and fractional chromarography would be ideal. Less


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Imagine a cube 1x1x1, then imagine that you form a cube of 10x10x10 with all these little cubes of 1x1x1. How many cubes do you have to remove to get rid of the surface of cubes.

4 Answers

10*10*2 + 10*8*2 + 8*8*2


n^3 - x = (n-2)^3 x = n^3 - (n-2)^3 x = 6n^2 - 12n +8 if n = 10 then x = 488 surface cubes to remove Less

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Canopy Growth Corporation

How do you deal with a difficult situation/conflict?

4 Answers

... up in hopes of progress towards resolution. There is always plan “B”😀

How do I know if I got the job? Lol

Actually, Electro/Mechanical is really my bag. I worked at Mitel and Zarlink Semiconductor for 16 years. I’ve Installed, configured, repaired and improved a multitude of equipment there and everywhere. If you have any need for a individual with a broad skill set I’m living near Smiths Falls and looking for a change. Thanks Less

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Mini Case: eBay is getting complaints from customers that the site doesn't have enough product selection. To increase product selection eBay runs a promotion and decides to waiver the listing fees for all sellers. eBay's earns revenue from the listing fees and as a % of sales transaction and the bulk of revenue comes as a commission over sales transactions. The promotion instead of increasing the revenue results in a drop in revenue for eBay. Explain

4 Answers

Ebay has actually tried this a few times. The listing fee keeps auctions that have no chance of selling off the site. If there is no listing fee, ebay loses that revenue and still only sees the same amount of auctions close with a sale. Buyers also tire of looking through a glut of items up for sale that they aren't interested in. Less

I asked a few questions about what percent of revenue comes from each of the revenue stream. Came up with a few recommendations on running a more focused promotion campaign to incentive sellers with the right kind of products. This would ensure that promotion budget is optimized and also help buyers see only the kind of products that they were interested in purchasing. Less

Let's keep this equation, Revenue = Listing fee + % commision * (orders * Avg order value) i'm assuming the hypothesis of removing the listing fee is to enable more products are introduced into the market and hence likely more orders. This could lead to the following outcome the # of orders are going up, but the AOV is dropping - It's possible that people are starting to sell items which they might have not before, because the listing fee eats into their margins. Less

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