If you use a 0.05 level of significance in a two-tail hypothesis test, what decision will you make if ZSTAT = +1.2
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If you use a 0.05 level of significance in a two-tail hypothesis test, what decision will you make if ZSTAT = +1.2
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- An environmentalist wishes to conduct a hypothesis test on the percentage of cars driven in the city that are hybrids. Is it sufficient for him to use a simple random sample of 173 cars if hybrids currently account for 8%of the car sales in the country and he claims that the percentage of hybrids in the city is higher than that?You are conducting a study to see if the proportion of voters who prefer the Democratic candidate is significantly larger than 50% at a level of significance of a = 0.10. According to your sample, 49 out of 92 potential voters prefer the Democratic candidate.US Universities found that 72% of people are concerned about the possibility that their personal records could be stolen over the internet. If a random sample of 300 college students at a Midwestern university were taken and 228 of them were concerned about the possibility that their personal records could be stolen over the Internet, could you conclude at the 0.025 level of significance that a higher proportion of the university’s college students are concerned about Internet theft than the public at large? Report the p-value for this test. Z0.025 = 1.96
- The business college wants to determine the proportion of business students who have extended time between classes. If the proportion differs from 30%, then the lab will modify a proposed enlargement of its facilities. Suppose a hypothesis test is conducted and the test statistic is 2.5. Find the P-value for a two-tailed test of hypothesis.A local coach wants to see if training in light rooms or dark rooms impacts athlete performance. With the single team, the coach tries training in different conditions and takes measurements. She ran her analysis in Microsoft Excel and received the following output table. If the P value is less than 0.05, the null hypothesis may be rejected. Based on the data table, what decision should the coach make about the hypothesis? Coach should reject the null hypothesis. Coach should fail to reject the null hypothesis. Coach should accept the null hypothesis. Coach does not have enough information to decide.Spam: A researcher reported that 71.8 % of all email sent in a recent month was spam. A system manager at a large corporation believes that the percentage at his company may be 69%. He examines a random sample of 500 emails received at an email server, and finds that 365 of the messages are spam. Can you conclude that greater than 69% of emails are spam? Use both a=0.01 and a= 0.05 levels of significance and the critical value method with the table. Part 1 of 5 State the appropriate null and alternate hypotheses. Ho: P- .69 H1: p> .69 This hypothesis test is a right-tailed V test. Part 2 of 5 Find the critical values. Round the answers to three decimal places. For a=0.01 , the critical value is 2.326 For a=0.05 , the critical value is 1.645 Part: 2/5 Part 3 of 5 Compute the test statistic. Do not round intermediate calculations. Round the answer to two decimal places.
- Management of a theater in a large city wants to know whether its popularity increased after it ran a popular musical. According to a previous study, 25% of city residents said they had been to the theater in the past year. A more recent study, conducted after the musical had run for a year, found that 156 of 536 residents sampled said they had been to the theater in the past year. Is there evidence at the ? = 0.05 level that the number of residents who have been to the theater has increased in the last year?According to a book published in 2011, 45% of the undergraduate students in the United States show almost no gain in learning in their first two years of college (Richard Arum et al., Academically Adrift, University of Chicago Press, Chicago, 2011). A recent sample of 1540 undergraduate students showed that this percentage is 37%. Can you reject the null hypothesis at a 2.5% significance level in favor of the alternative that the percentage of undergraduate students in the United States who show almost no gain in learning in their first two years of college is currently lower than 45%. Use both the p-value and the critical-value approaches. Round your answers for the observed value of z and the critical value of z to two decimal places, and the p-value to four decimal places. Zobserved = p-value = Critical value = i Hence we can conclude that the percentage of undergraduate students in the U.S. who show almost no gain in learning in their fırst two years of college is currently 45%.One study claimed that 88 % of college students identify themselves as procrastinators. A professor believes that the claim regarding college students is too high. The professor conducts a simple random sample of 272 college students and finds that 231 of them identify themselves as procrastinators. Does this evidence support the professor's claim that fewer than 88 % of college students are procrastinators? Use a 0.02 level of significance. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho p = 0.88 Ha P 0.88
- A two-tailed hypothesis test is being used to evaluate a treatment effect with α = .05. If the sample data produce a z-score of z = -2.24, which is the correct decision?An experiment was conducted to test if 2 new anti-anxiety medications will have an effect on number of anxiety symptoms. The researcher randomly assigned participants to one of the three experimental groups. Groups 1 and 2 took the 2 new medications while Group 3 took a placebo pill. The researcher was interested in determining whether there would be a difference in the number of anxiety symptoms among the three groups. Assume an alpha level of 0.05 and a two-tailed test. What statistical test should you use to evaluate the hypothesis above? one-way independent ANOVA Pearson Product-Moment Correlation simple linear regression multiple linear regression three-way independent ANOVAA dowser has correctly located water for a well 1 out of 2 times in Jones County. In Jones County, someone who is just guessing has a 40% chance of locating water for a well. Does this sample provide sufficient evidence that the dowser can locate water and is not just guessing?