A newspaper published an article about a study in which researchers subjected laboratory gloves to stress. Among 224 vinyl gloves, 60% leaked viruses. Among 224 latex gloves, 8% leaked viruses. Using the accompanying display of the technology results, and using a 0.05significance level, test the claim that vinyl gloves have a greater virus leak rate than latex gloves. Let vinyl gloves be population 1.i Click the icon to view the technology results.What are the null and alternative hypotheses?OA. Ho: P₁ P2H₁: P₁ = P2O D. Ho: P₁ = P₂H₁: P₁ P2Identify the test statistic.(Round to two decimal places as needed.)Identify the P-value.(Round to three decimal places as needed.)What is the conclusion for this test?The P-value isthe significance level a, soO B. Ho: P₁H₁: P₁OE. Ho: P₁the null hypothesis. There isP₂P2P₂H₁: P₁ = P2Technology Resultsevidence to support the claim that vinyl gloves have a greater virus leak rate than latex gloves.Pooled proportion: 0.34Test statistic, z: 11.5389Critical z: 1.6449P-value: 0.0000OC. Ho: P₁ P2H₁: P₁ P2- X

According to the given information in this question, we need to find the test statistic and p value…

Answered: are the null and alternative…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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The commercial for the new Meat Man Barbecue claims that it takes 14 minutes for assembly. A consumer advocate thinks that the assembly time is lower than 14 minutes. The advocate surveyed 12 randomly selected people who purchased the Meat Man Barbecue and found that their average time was 13.5 minutes. The standard deviation for this survey group was 2.1 minutes. What can be concluded at the the a = 0.01 level of significance level of significance? You may assume the data are normally distributed.
The null hypothesis and the alternate hypothesis are: H0: The frequencies are equal.H1: The frequencies are not equal. Category f0 A 10 B 10 C 20 D 10 a. State the decision rule, using the 0.01 significance level. (Round your answer to 3 decimal places.) b. Compute the value of chi-square. (Round your answer to 2 decimal place.) c. What is your decision regarding H0?
The average wait time to get seated at a popular restaurant in the city on a Friday night is 13 minutes. Is the mean wait time less for men who wear a tie? Wait times for 16 randomly selected men who were wearing a tie are shown below. Assume that the distribution of the population is normal. 11, 14, 14, 14, 13, 13, 10, 11, 10, 10, 12, 10, 12, 14, 13, 11 What can be concluded at the the a = 0.05 level of significance level of significance?
Consider testing the hypotheses Ho: p 0.4 versus Hai p> 0.4. Four possible sample statistics, along with four possible p-values, are given. Match the statistics to their p-values. p= 0.42 p=0.38 p= 0.51 p= 0.46 D
The average wait time to get seated at a popular restaurant in the city on a Friday night is 13 minutes. Is the mean wait time greater for men who wear a tie? Wait times for 11 randomly selected men who were wearing a tie are shown below. Assume that the distribution of the population is normal. 12, 13, 13, 15, 13, 11, 14, 13, 15, 11, 15 What can be concluded at the the a = 0.01 level of significance level of significance?
The average fruit fly will lay 394 eggs into rotting fruit. A biologist wants to see if the average will change for flies that have a certain gene modified. The data below shows the number of eggs that were laid into rotting fruit by several fruit flies that had this gene modified. Assume that the distribution of the population is normal. Round your mean and standard deviation to three decimal places when doing these calculations. 376, 400, 402, 362, 386, 400, 368, 391, 399, 407, 384, 384, 371, 371 Download CSV What can be concluded at the the a = 0.01 level of significance level of significance?
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The average American consumes 85 liters of alcohol per year. Does the average college student consume less alcohol per year? A researcher surveyed 13 randomly selected college students and found that they averaged 82.1 liters of alcohol consumed per year with a standard deviation of 23 liters. What can be concluded at the the a = 0.10 level of significance? You may assume the data are normally distributed.
Before the furniture store began its ad campaign, it averaged 222 customers per day. The manager is investigating if the average is smaller since the ad came out. The data for the 14 randomly selected days since the ad campaign began is shown below. Round your mean and standard deviation to three decimal places when doing these calculations. 224, 211, 206, 206, 232, 189, 222, 197, 220, 210, 212, 213, 208, 189 Download CSV Assuming that the distribution is normal, what can be concluded at the a = 0.01 level of significance?
The data below show sport preference and age of participant from a random sample of members of a sports club. Is there evidence to suggest that they are related? Frequencies of Sport Preference and Age Swimming 85 ATTI Tennis Basketball 18-25 98 88 26-30 104 100 46 31-40 62 63 40 54 30 Over 40 63 What can be concluded at the = 0.10 significance level? a. What is the correct statistical test to use? O Homogeneity O Paired t-test O Goodness-of-Fit O Independence b. What are the null and alternative hypotheses? Ho: O The age distribution is the same for each sport. The age distribution is not the same for each sport. O Age and sport preference are dependent. O Age and sport preference are independent. H1: O Age and sport preference are independent. O The age distribution is not the same for each sport. O Age and sport preference are dependent. O The age distribution is the same for each sport.
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The null hypothesis and the alternate hypothesis are: H: The frequencies are equal. H₁: The frequencies are not equal. Category fe ABC D 15 10 10 15 a. State the decision rule, using the 0.05 significance level. (Round your answer to 3 decimal places.) Reject Ho if chi-square > b. Compute the value of chi-square. (Round your answer to 2 decimal place.) Chi-square value

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