1. Lighthouse

Level 4 · Test the test with an A/A split

The request

From Priya Shah, Head of Data

“Before we trust any A/B result, I want proof our method doesn't invent winners. Split all sessions into two halves that saw the same site, even and odd session ids, and test them against each other. What do z and p come out to?”

Your goal

Arm B = sessions with an even session_id, arm A = odd. Set result to a dict with "z": (rate B − rate A) / se, where se = √(p × (1 − p) × (1/n_A + 1/n_B)) and p is the pooled rate (all conversions / all sessions), rounded to 3 decimals; and "p_value": the two-sided p-value 2 × (1 − Φ(|z|)) from the unrounded z, rounded to 4 decimals.

  • Use the web_sessions DataFramenot done yet
  • Store a dict in resultnot done yet

Your code

Python

Loading the editor…

    ResultPreview of web_sessions · press Run to see yours
    Preview of web_sessions
    session_idcustomer_idstarted_atdevicesource
    1NaN2023-01-01 23:02:24mobilepaid
    2182023-01-01 15:07:49desktoppaid
    3NaN2023-01-01 16:55:46desktopdirect
    4NaN2023-01-01 12:43:28mobileorganic

    Mote checks your result

    Press Run. Each row that matches the goal becomes a firefly in your lantern.

    DataFrames

    Loaded for you as pandas DataFrames, plus pd and np.

    web_sessions40,000 rows · 7 columns
    • session_idinteger
    • customer_idinteger
    • started_attimestamp
    • devicevarchar
    • sourcevarchar
    • pages_viewedinteger
    • convertedboolean
    First 5 rows of web_sessions
    session_idcustomer_idstarted_atdevicesourcepages_viewedconverted
    1NaN2023-01-01 23:02:24mobilepaid1false
    2182023-01-01 15:07:49desktoppaid1false
    3NaN2023-01-01 16:55:46desktopdirect7false
    4NaN2023-01-01 12:43:28mobileorganic3false
    5NaN2023-01-01 13:06:21mobilesocial2false
    customers2,018 rows · 9 columns
    • customer_idinteger
    • first_namevarchar
    • last_namevarchar
    • emailvarchar
    • cityvarchar
    • statevarchar
    • signup_datedate
    • segmentvarchar
    • referral_sourcevarchar
    First 5 rows of customers
    customer_idfirst_namelast_nameemailcitystatesignup_datesegmentreferral_source
    1OwenHudsonowen.hudson@icloud.comSan FranciscoCA2022-06-02consumersearch
    2DouglasMendozadmendoza69@gmail.comAtlantaGA2022-06-02consumerNaN
    3HenryBrookshbrooks@lakesidecafe.comLos AngelesCA2022-06-04businessNaN
    4BrianGrahambgraham@bluecedarhomes.comSeattleWA2022-06-06businesspinterest
    5KyleJimenezkjimenez27@yahoo.comIndianapolisIN2022-06-07consumerNaN
    products120 rows · 8 columns
    • product_idinteger
    • namevarchar
    • categoryvarchar
    • subcategoryvarchar
    • unit_pricedecimal
    • unit_costdecimal
    • launched_ondate
    • is_activeboolean
    First 5 rows of products
    product_idnamecategorysubcategoryunit_priceunit_costlaunched_onis_active
    101Vale Velvet LoveseatFurnitureSofas1659.00NaN2023-09-11true
    102Alder Oak Coffee TableFurnitureTables919.00503.502024-05-16true
    103Orchard Rattan Dining ChairFurnitureChairs789.00429.872023-05-27true
    104Upland Oak BookcaseFurnitureStorage429.00214.702022-12-20true
    105Avery Leather Bed FrameFurnitureBeds449.00232.702021-08-31true
    orders12,000 rows · 7 columns
    • order_idinteger
    • customer_idinteger
    • order_datetimestamp
    • statusvarchar
    • channelvarchar
    • shipping_feedecimal
    • coupon_codevarchar
    First 5 rows of orders
    order_idcustomer_idorder_datestatuschannelshipping_feecoupon_code
    10011252023-01-01 20:51:19completedweb0.00NaN
    10021562023-01-02 07:58:46completedweb0.00NaN
    1003222023-01-02 10:25:27completedmarketplace4.99WELCOME10
    1004482023-01-02 11:21:51completedmarketplace4.99WELCOME10
    1005822023-01-02 11:43:36completedweb0.00WELCOME10
    order_items30,338 rows · 6 columns
    • order_item_idinteger
    • order_idinteger
    • product_idinteger
    • quantityinteger
    • unit_pricedecimal
    • discount_pctdecimal
    First 5 rows of order_items
    order_item_idorder_idproduct_idquantityunit_pricediscount_pct
    11001121931.990.00
    21001154311.990.00
    31002154411.990.00
    410021941149.000.00
    51002193219.990.00
    returns1,673 rows · 5 columns
    • return_idinteger
    • order_item_idinteger
    • returned_ondate
    • reasonvarchar
    • refund_amountdecimal
    First 5 rows of returns
    return_idorder_item_idreturned_onreasonrefund_amount
    1142023-01-10damaged83.69
    2422023-01-10wrong_item19.99
    31052023-01-14wrong_item10.99
    4242023-01-16changed_mind26.99
    51262023-01-16late20.99
    employees60 rows · 8 columns
    • employee_idinteger
    • first_namevarchar
    • last_namevarchar
    • departmentvarchar
    • titlevarchar
    • manager_idinteger
    • hire_datedate
    • salaryinteger
    First 5 rows of employees
    employee_idfirst_namelast_namedepartmenttitlemanager_idhire_datesalary
    1CatherineHaleExecutiveChief Executive OfficerNaN2016-03-01285000
    2RobertKimExecutiveChief Operating Officer12017-01-09235000
    3MeganCollinsSalesHead of Sales12018-04-16172000
    4ChloeGutierrezSalesAccount Executive32020-01-2172000
    5FrankBellSalesAccount Executive32023-05-2173000
    marketing_spend3,276 rows · 3 columns
    • spend_datedate
    • channelvarchar
    • amountdecimal
    First 5 rows of marketing_spend
    spend_datechannelamount
    2023-01-01paid492.99
    2023-01-01email37.38
    2023-01-01social200.53
    2023-01-02paid709.13
    2023-01-02email64.13

    The lesson

    Two rates, one question

    An A/B test ends with two conversion rates. The question is whether the gap between them is bigger than the gap you would expect from luck alone. The two-proportion z-test answers it by measuring the gap in standard errors:

    z = (rate_B - rate_A) / se

    A z of 0 means no gap. A z around ±2 means the gap is about two standard errors wide, which happens by chance only about 5% of the time.

    The pooled standard error

    The test assumes the null hypothesis (no difference), so it estimates one shared rate from both arms together, the pooled rate, and uses it for the standard error of the difference:

    pooled = (conv_a + conv_b) / (n_a + n_b)
    se = math.sqrt(pooled * (1 - pooled) * (1 / n_a + 1 / n_b))

    On the A/A split each arm has 20,000 sessions. Arm A converts 577 times (2.89%), arm B 616 times (3.08%). The pooled rate is 2.98% and the standard error of the difference about 0.17 points.

    From z to a p-value

    Python’s standard library has the normal curve built in. NormalDist().cdf(z) is the share of the curve below z, so the chance of a gap at least this extreme in either direction is:

    from statistics import NormalDist
    p_value = 2 * (1 - NormalDist().cdf(abs(z)))

    Here z ≈ 1.15 and p ≈ 0.25: a gap like this between two identical halves shows up in a quarter of splits. Nothing to see, which is exactly right, because nothing was different.

    Why run an A/A test

    An A/A test checks the machinery: assignment, logging, the analysis code. Over many A/A tests about 5% should come out with p < 0.05, because that is what the 0.05 threshold means. It really happens: take the same even/odd split for 2024 only and the full year comes out at p = 0.047. Two halves of identical traffic, “significant”. That is not a bug, it is the 1-in-20 the threshold accepts, and the reason one significant test is evidence, not proof.

    Common mistakes

    • One-sided by accident. Leaving out the 2 × halves the p-value. Use two-sided unless you decided the direction before the test started.
    • Unpooled se in the test. Each arm’s own rate is right for a confidence interval of the difference; the pooled rate is the standard choice for the test itself.
    • Rounding z before computing p. Round only the reported numbers.
    • Sessions that aren’t independent. If one shopper’s visits land in both arms, the test is overconfident. Split by customer when you can.

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