# G correlation hypothesis test two different

• Test Prep
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g) Correlation Hypothesis Test Two different quantitative variables (ordered pair) Ho: rho = 0 (no correlation) Ha: rho not = 0 (is correlation) Assumptions for Correlation Rho Test: Random ordered paired data , sample sizes at least 30, Scatterplot shows a linear shape (No pattern in the scatterplot or in the verses fit residual plot), Residuals are nearly normal, Residuals are centered close to zero, No fan shape in the residual plot vs x value (Homoscedasticity), No outliers that are overly influential T-Test statistic: The number of standard errors that the slope of the regression line is above or below zero. 2. P1: percent of babies exposed to cocaine that passed the test P2: percent of babies not exposed to cocaine that passed the test H0: P1 = p2 HA: P1 < p2 (claim) Assumptions: Passed the assumptions. The data was random, There was at least 10 successes and failures in both groups. ( x1 = 139 > 10 , n1-x1 = 51 > 10 , x2 = 153 > 10 , n2 x2 = 33 > 10 ) Test Statistic: z = -2.12 . This means that the sample percent of cocaine babies is 2.12 standard deviations below the sample percent of babies not exposed to cocaine. There is a significant difference between the groups. P value = 0.0164. This means that if the null hypothesis is true and the percent of children exposed to cocaine is the same as the percent of children not exposed, then there was a 1.64% chance of getting the sample values or more extreme. It is very unlikely that this data would happen by random chance from equal populations. Reject H0. Conclusion: There is sufficient evidence to support the claim that the passage rates

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for babies that were exposed to cocaine is lower than the passage rates for babies not exposed to cocaine on the test of object assembly. 3. 1 2 : Ave weight of male German Shepherds : Ave weight of male Dobermans The problem does meet the assumptions. Both data sets were random and despite the data sets being too small (n1=20 and n2=14) , they were normally distributed. The data sets are independent. There is no relationship between the weight of a random German Shepherd and the weight of a random Doberman. (Unless they have the same owner which was very unlikely.) We will perform a regular 2 pop T-test. (Not paired) 0 1 2 2 : : (claim) A H H Test Stat t = 0.558 This means that the sample mean weight of the German Shepherds in the data is only 0.558 standard errors above the sample mean weight of the Dobermans. There was not a significant difference between the sample means. They were very close. P-value = 0.2906 . This means that if the null hypothesis is true and the mean average weight of German Shepherds and Dobermans are the same, then there was a 29.06% chance of getting the sample values or more extreme. This data could of happen by random chance from equal population means.
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• Fall '17
• Statistics, Statistical hypothesis testing, significant sample evidence

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