statanalysis.conf_inte_md package

Submodules

todo - refactor output (last lines) - use “alternative” instead of “tail” - use kwargs format while calling functions - reorder fcts attributes

statanalysis.conf_inte_md.ci_estimators.CIE_MEAN_ONE(n, mean_dist, std_sample, t_etoile=None, cf: float | None = None)

Get_interval_mean:get the mean of a population from a sample (no sign pb) - cf: confidence level (or coverage_probability) - n: number of observations == len(sample) - mean_dist: the mean measured on the sample = mean(sample) - std_sample: std of the sample ==std(sample) - t_etoile: if set, cf is ignored.

Hyp - better the population follow nornal dist. Or use large sample (>10)

  • Alternative to normality: Wilcoxon Signed Rank Test

Theo - reade [here](https://en.wikipedia.org/wiki/Student’s_t-distribution#How_Student’s_distribution_arises_from_sampling)

statanalysis.conf_inte_md.ci_estimators.CIE_MEAN_TWO(N1, N2, diff_mean, std_sample_1, std_sample_2, t_etoile=None, pool=False, cf: float | None = None)

Get_interval_diff_mean: get the diff in mean of two populations(taking their samples) (sign(diff_mean) => no sign pb) - cf: confidence level (or coverage_probability) - N1: number of observations == len(sample1) - N2: number of observations == len(sample2) - mean_dist: the mean measured on the sample = mean(sample) - std_sample_1: std of the sample ==std(sample1) - std_sample_2: std of the sample ==std(sample2) - t_etoile: if set, cf is ignored. - pool: default False

  • True

    • if we assume that our populations variance are equal

    • we use a t-distribution of (N1+N2-1) ddl

  • False

    • if we assume that our populations variance are not equal

    • we use a t-distribution of min(N1, N2)-1 ddl

Hyp - both the population follow normal dist. Or use large sample (>10) - the populations are independant from each other - use simple random samples - for pool=True, variances are assume to be the same

  • to test that, you can
    • use levene test [plus robuste que fusher ou bartlett face à la non-normalité de la donnée](https://fr.wikipedia.org/wiki/Test_de_Bartlett)

      • H0: Variances are equals; H1: there are not

      `python scipy.stats.levene(liste1,liste2, center='mean') solution = "no equality" if p-value<0.05 else "equality" `

    • or check if IQR are the same

      • IQR = quantile(75%) - quantile(25%)

Eqvl - scipy.stats.ttest_ind(liste1,liste2, equal_var = False | True)

Eqvl_pointWise estimation - Assume diff_mean = 82 - Result: diff_mean in CI = [77.33, 87.63] - If we test H0:p=80 vs H1:p>80, we would fail to reject the null because H1 is not valide here - As sa matter of fact, there is some value in CI below 80 witch if not compatible with H1 => the test doest give enough evidence to reject H0

Theo - read [here](https://en.wikipedia.org/wiki/Student’s_t-distribution#How_Student’s_distribution_arises_from_sampling)

statanalysis.conf_inte_md.ci_estimators.CIE_ONE_PROPORTION(proportion, n, method, cf: float | None = None)

Get_interval_simple: get a proportion of an attribute value (male gender, ) in a population based on a sample (no sign pb) - cf: confidence_level (or coverage_probability) - proportion: measurement - n: number of observations == len(sample) - method: “classic” or “conservative”

Hyp - better the population follow nornal dist. Or use large sample (>10)

statanalysis.conf_inte_md.ci_estimators.CIE_PROPORTION_TWO(p1, p2, n1, n2, cf: float | None = None)

Get_interval_diff: get the diff of mean between 2 population based on a sample from each population (p1-p2) #p1-p2# - cf: confidence_level (or coverage_probability) - p1: mean of liste1 - p2: mean of liste2 - n1: len(liste1) - n2: len(liste2)

Hyp - better the populations follow normal dist. Or use large samples (>10)

statanalysis.conf_inte_md.ci_estimators.get_min_sample(moe: float, p=None, method=None, cf: float | None = None)

Get_min_sample:get the minimum of sample_size to use for a - input

  • cf: confidence (or coverage_probability): between 0 and 1

  • moe: margin of error

  • method (optional): “conservative” (default)

  • p: not used if method==”conservative”

Hyp - better the population follow nornal dist. Or use large sample (>10)

statanalysis.conf_inte_md.ci_estimators.print(*args)

Some defs - parameter: A quantifiable characteristic of a population - confidence interval: range of reasonable values for the parameter

todo - use kwargs format while calling functions - reorder fcts attributes

statanalysis.conf_inte_md.confidence_interval.IC_MEAN_ONE(sample: list, t_etoile=None, confidence: float | None = None)

Estimate_population_mean(ONE MEAN): We need the spread (std): We will use an estimation

Data

  • confidence:..

  • sample: value…

Method - Use t-distribution to calculate few

Hypothesis - Samples follow a normal (or large enough to bypass this assumption) => means of these sample follow a t-dist

statanalysis.conf_inte_md.confidence_interval.IC_MEAN_TWO_NOTPAIR(sample1, sample2, pool=False, confidence: float | None = None)

Difference_population_means_for_nonpaired_data(TWO MEANS FOR PAIRED DATA): We have have estimate a parameter p on two populations (1 , 2).How to estimate p1-p2 ? #p1-p2#

Construction - It is like,

  • checking if a feature magnitude change when going from a category to another

  • Example_contexte
    • having a dataframe df, with 3 col [name, score, equipe, role]

      • equipe: “1” or “2”

      • role: df.role.nunique = 11 => len(df)==22

    • Now there is a battle: For a “same role” fight”, which team is the best?

  • Example_question
    • if education level are generally equal -> mean difference is 0

      • Is there a mean difference between the education level based on gender

      • if education levels are unequel -> mean difference is not 0

    • So, Look for 0 in the ranfe of reaonable values

We need the spread (std): We will use an estimation

Args

  • confidence:..

  • Sample1: list: values…

  • Sample2: list: (same len) values…

  • pool: default False
    • True

      • if we assume that our populations variance are equal

      • we use a t-distribution of (N1+N2-1) ddl

    • False

      • if we assume that our populations variance are not equal

      • we use a t-distribution of min(N1, N2)-1 ddl

Method - Use t-distribution to calculate few - create joint confidence interval

Hypothesis - a random sample - Samples follow a normal (or large enough to bypass this assumption: 10 per category) => means of these sample follow a t-dist

Notes - With {cf} confidence, the population mean difference of the (second_team - first_team) attribute is estimated to be between {data.interval[0]} and {dat.interval[1]} - if all values are above 0, cool there is a significativity

statanalysis.conf_inte_md.confidence_interval.IC_MEAN_TWO_PAIR(sample1, sample2, t_etoile=None, confidence: float | None = None)

Difference_population_means_for_paired_data(TWO MEANS FOR PAIRED DATA): We have have estimate a parameter p on two populations (1 , 2).How to estimate p1-p2 ? #p1-p2#

What is paired data ?

  • measurements took on individuals (people, home, any object)

  • technicality:

    • When in a dataset (len = n) there is a row df.a witch values only repeat twice (=> df.a.nunique = n/2)

    • we can do a plot(x=feature1, y=feature2)

  • examples

    • Each home need canibet quote from two suppliers => we want to know if there is an average difference in nb_quotes from between twese two suppliers

    • In a blind taste test to compare two new juice flavors, grape and apple, consumers were given a sample of each flavor and the results will be used to estimate the percentage of all such consumers who prefer the grape flavor to the apple flavor.

  • Construction
    • It is like,

      • checking if a feature magnitude change when going from a category to another, each pair split the two cztegories

      • Example_contexte
        • having a dataframe df, with 3 col [name, score, equipe, role]

          • equipe: “1” or “2”

          • role: df.role.nunique = 11 => len(df)==22

        • Now there is a battle: For a “same role” fight”, which team is the best?

      • Example_question
        • if education level are generally equal -> mean difference is 0

          • Is there a mean difference between the education level of twins

          • if education levels are unequel -> mean difference is not 0

        • So, Look for 0 in the ranfe of reaonable values

We need the spread (std): We will use an estimation

Equivl - IC_MEAN_ONE(confidence, sample1 - sample2)

Data

  • confidence:..

  • Sample1: list: values…

  • Sample2: list: (same len) values…

Method - Use t-distribution to calculate few - create joint confidence interval

Hypothesis - a random sample of identical twin sets - Samples follow a normal (or large enough to bypass this assumption: (ex 20 twins)) => means of these sample follow a t-dist

Notes - With {cf} confidence, the population mean difference of the (second_team - first_team) attribute is estimated to be between {data.interval[0]} and {dat.interval[1]} - if all values are above 0, cool there is a significativity

statanalysis.conf_inte_md.confidence_interval.IC_PROPORTION_ONE(sample_size: int, parameter: float, confidence: float | None = None, method: str | None = None)

Confidence_interval(ONE PROPORTION):Confidence interval calculus after a statistic test - input

  • sample_size: int: sample size (more than 10 to use this method)

  • parameter: float: the measurement on the sample

  • confidence: float: confidence confidence (between O and 1). Greater the confidence, wider the interval

  • method: str: either “classic” (default) or “conservative.

  • Example:

    • how many men in the entire population with a con ?

    • a form filled by 300 people show that there is only 120 men => p = (120/300); N=300

  • Hypothesis

    • the sample is over 10 for each of the categories in place => we use the “Law of Large Numbers”

    • the sample proportion comes from data that is considered a simple random sample

  • Idea

    • let P: the real proportion in the population

    • let S: Size of each sample == nb of observations per sample

    • For many samples, we calculate proportions per sample: ex: for N samples of size S => N proportions values

    • (p - P) / ( p*(1-p)/S ) follow a normal distribution

  • Descriptions:

    • For a given polulation and a parameter P to find, If we repeated this study many times, each producing a new sample (of the same size {res.sample_size==S}) from witch a {res.confidence} confidence interval is computed, then {res.confidence} of the resulting confidence intervals would be excpected to contain the true value P

    • If the entire interval verify a property, then it is reasonable say that the parameter verify that property

  • Result

    • with a {res.confidence} confidence, we estimate that the populztion proportion who are men is between {res.left_tail} and {res.right_tail}

statanalysis.conf_inte_md.confidence_interval.IC_PROPORTION_TWO(p1, p2, N1, N2, confidence: float | None = None)

Difference_population_proportion(TWO PROPORTIONS): We have have estimate a parameter p on two populations (1 , 2).How to estimate p1-p2 ? #p1-p2#

Method

  • create joint confidence interval

Construction - Cmparison

Hypotheses - two independant random samples - large enough sample sizes : 10 per category (ex 10 yes, 10 no)

statanalysis.conf_inte_md.confidence_interval.print(*args)

Module contents