Hypothesis Testing 1S2T - Call Center Process. Sample Parameters: n=50, df=50-1=49, Mean1=4, SD1=3 1-sample 2-tail ttest Assume Null Hypothesis Ho as Mean1 = 4 Thus, Alternate Hypothesis Ha as Mean1 ≠ 4

Data Science - Hypothesis Testing Work

Hypothesis-Testing-1-Sample-1-Tail-Test-Salmonella-Outbreak. 1-sample 1-tail ttest. Assume Null Hypothesis Ho as Mean Salmonella <= 0.3. Thus Alternate Hypothesis Ha as Mean Salmonella > 0.3. As No direct code for 1-sample 1-tail ttest available with unknown SD and arrays of means. Hence we find probability using 1-sample 2-tail ttest and divide…

Testing the hypothesis of a pharma company for the time of the effect and the quality assurance based on the null and alternate hypothesis

ANOVA test using python to find out if survey or experiment results are significant and the impact of one or more factors by comparing the means of different samples

Hypothesis-Testing-2-Proportion-T-test-Students-Jobs-in-2-States. Assume Null Hypothesis as Ho is p1-p2 = 0 i.e. p1 ≠ p2. Thus Alternate Hypthesis as Ha is p1 = p2. Explanation of bernoulli Binomial RV: np.random.binomial(n=1,p,size) Suppose you perform an experiment with two possible outcomes: either success or failure. Success happens with pro…

Chi2 contengency independence test Q4. TeleCall uses 4 centers around the globe to process customer order forms. They audit a certain % of the customer order forms. Any error in order form renders it defective and has to be reworked before processing. The manager wants to check whether the defective % varies by centre. Please analyze the data at 5

Hypothesis-Testing-Chi2-Test-Human-Gender-and-Choice-of-Pets. Assume Null Hypothesis as Ho: Human Gender and choice of pets is independent and not related. Thus Alternate Hypothesis as Ha : Human Gender and choice of pets is dependent and related. As (p_valu=0.1031) > (α = 0.05); Accept Null Hypothesis i.e Independence among categorical variable…

Chi2 contengency independence test Q5. Fantaloons Sales managers commented that % of males versus females walking in to the store differ based on day of the week. Analyze the data and determine whether there is evidence at 5 % significance level to support this hypothesis. Assume Null Hypothesis as Ho: Independence of categorical variables (% of

Hypothesis-Testing-2-Sample-2-Tail-Test-Drugs-and-Placebos. Note: This python code states both 2-sample 1-tail and 2-sample 2-tail codes. Treatment group mean is Mu1 Contrl group mean is Mu2 2-sample 2-tail ttest Assume Null Hypothesis Ho as Mu1 = Mu2 Thus Alternate Hypothesis Ha as Mu1 ≠ Mu2.

Hypothesis Testing Anova Test - Iris Flower dataset. Anova ftest statistics: Analysis of varaince between more than 2 samples or columns. Assume Null Hypothesis Ho as No Varaince: All samples population means are same. Thus Alternate Hypothesis Ha as It has Variance: Atleast one population mean is different. As (p_value = 0) < (α = 0.05); Reject…

Chi2 contengency independence test. Fantaloons Sales managers commented that % of males versus females walking in to the store differ based on day of the week. Analyze the data and determine whether there is evidence at 5 % significance level to support this hypothesis.

Analyze ab test results udacity project

Used libraries and functions as follows:

Chi2 contengency independence test. Q4. TeleCall uses 4 centers around the globe to process customer order forms. They audit a certain % of the customer order forms. Any error in order form renders it defective and has to be reworked before processing. The manager wants to check whether the defective % varies by centre. Please analyze the data a…

Hypothesis-Testing-Chi2-Test-Athletes-and-Smokers. Assume Null Hypothesis as Ho: Independence of categorical variables (Athlete and Smoking not related). Thus Alternate Hypothesis as Ha: Dependence of categorical variables (Athlete and Smoking is somewhat/significantly related). As (p_value = 0.00038) < (α = 0.05); Reject Null Hypothesis i.e. De…