### Understanding Hypothesis Testing for Glove Virus Leak RatesA newspaper reported a study in which researchers tested laboratory gloves for virus leakage. The study involved 258 vinyl gloves, with a 67% leak rate, and 258 latex gloves, with an 8% leak rate. The aim is to test, at a 0.01 significance level, whether vinyl gloves have a higher virus leak rate than latex gloves. Vinyl gloves are designated as population 1.#### HypothesesThe task is to determine the null and alternative hypotheses:- **Option B** is selected: - Null Hypothesis (H₀): \( p_1 = p_2 \) - Alternative Hypothesis (H₁): \( p_1 > p_2 \)This implies the hypothesis test is set to check if the proportion of leaks in vinyl gloves (\( p_1 \)) is greater than in latex gloves (\( p_2 \)).#### How to Identify the Test StatisticThe test statistic should be calculated to examine the validity of the null hypothesis, rounding the result to two decimal places. You would typically apply a z-test for two proportions in this context, comparing vinyl and latex gloves' leak rates.By conducting this hypothesis testing, we can scientifically evaluate claims about the performance of different glove materials in preventing virus leakage.#### Visual and Interactive Elements- **Icon for Technology Results**: There is an icon that indicates the availability of further data, likely leading to detailed results or datasets necessary to compute the test statistic and perform the hypothesis test.This setup allows learners to understand hypothesis testing principles, apply statistical tests, and interpret results within practical scenarios. **Title: Statistical Analysis of Virus Leakage in Laboratory Gloves****Background:**In a study, researchers tested the susceptibility of laboratory gloves to virus stress. The study focused on two types of gloves: vinyl and latex. Among 258 vinyl gloves, 67% were found to leak viruses, whereas a lower rate of leakage was reported for latex gloves.**Objective:**The study aims to determine if vinyl gloves exhibit a significantly greater virus leak rate compared to latex gloves. Vinyl gloves are categorized as population 1 and latex gloves as population 2 for statistical analysis.**Hypotheses:**- Null Hypothesis (H₀): p₁ = p₂ (The proportion of leakage in vinyl gloves is equal to that in latex gloves.)- Alternative Hypothesis (H₁): p₁ > p₂ (The proportion of leakage in vinyl gloves is greater than that in latex gloves.)**Statistical Analysis:**A technology-assisted analysis was performed, producing the following key results:- Pooled Proportion: 0.38- Test Statistic (z): 13.8146- Critical Z: 2.3264- P-value: 0.0000- 98% Confidence Interval: 0.5103920 < p₁ < 0.6679026**Interpretation:**1. **Pooled Proportion:** The average proportion of leakage between both types of gloves.2. **Test Statistic (z):** A measure that indicates how many standard deviations the sample proportion is from the null hypothesis proportion.3. **Critical Z:** The threshold value for the z-statistic at a 98% confidence level.4. **P-value:** The probability that the observed data would occur if the null hypothesis were true. A p-value of 0.0000 suggests a very strong evidence against the null hypothesis.5. **Confidence Interval:** This interval estimates where the true proportion of virus leakage (p₁) in vinyl gloves likely lies, with 98% confidence.**Conclusion:**The analysis strongly supports the alternative hypothesis that vinyl gloves have a higher rate of virus leakage in comparison to latex gloves, given the test statistic and the p-value. The confidence interval reinforces this finding, indicating a higher proportion of leakage in vinyl gloves. This results in a recommendation for increased caution or selection against vinyl gloves when assessing virus penetration risks.

Given data in the problem isn1 = 258n2 = 258p1 = 0.67p2 = 0.08We have to calculate test statistics (a)Null hypothesis H0 : P1 = P2Alternative hypothesis H1 : P1 > P2(b) P = n1p1 + n2p2 / n1 + n2 = 258(0.67) + 258(0.08) / 258+258 = 172.86 + 20.64 / 516 = 193.5 / 516 P = 0.375 Q = 1 - P = 1 - 0.375Q = 0.625SE = √ PQ ( 1/n1 + 1/n2 )=√ 0.375 * 0.625 ( 1/ 258 + 1/258 ) =√ 0.234375 ( 0.003875968 + 0.003875968 ) =√ 0.234375(0.007751936) =√ 0.00181686SE = 0.042624 Test statistics Z = p1 - p2 / SEZ = 0.67 - 0.08 / 0.042624 = 0.59 / 0.042624Z = 13.8419Z ~ 13.8146Z ~ 13.81