mean credit score to be 721.3 with a standard deviation of 81.9. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the a = 0.05 level of significance. State the null and alternative hypotheses.
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- 10. Gravetter/Wallnau/Forzano, Essentials - Chapter 9 - End-of-chapter question 15 Weinstein, McDermott, and Roediger (2010) report that students who were given questions to be answered while studying new material had better scores when tested on the material compared to students who were simply given an opportunity to reread the material. In a similar study, an instructor in a large psychology class gave one group of students questions to be answered while studying for the final exam. The overall average for the exam was u = 73.4, but the n = 16 students who answered questions had a mean of M = 78.3 with a standard deviation of s = 8.4. For this study, did answering questions while studying produce significantly higher exam scores? Use a one-tailed test with a = .01 and the Distributions tool to help. (Round your answers to three decimal places, when needed.) t Distribution Degrees of Freedom = 21 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 SM %3D t-critical %3DA quiz related to digital knowledge was conducted. The quiz had 10 questions and covered topics such as the purpose of browser cookies, phishing scams, and privacy policies. The survey was given to 50 people. The mean score is 4.1 with a standard deviation of 2.6. We want to know if the data provide evidence that the mean score in the population is lower than 5. We will use alpha = 0.05 to make our decision. What is the null hypothesis for this question? Group of answer choices a. The proportion of people who pass the test is 5. b. The mean test score of the population is 5. c. The mean test score of the 50 people who took the test is 5. d.The standard deviation is 2.6As Director of Admissions at Garden Valley University, you believe that the average age of full-time students (Group 1) is less than the average age of part-time students (Group 2). You sample 45 full-time students, and the sample mean age is 22.3 with a standard deviation of 4.8. You sample 50 part-time students, and the sample mean age is 24.3 with a standard deviation of 5.6. Test the claim using a 2% level of significance. Assume the population standard deviations are unequal and that both populations are normally distributed. Where necessary, round results accurate to 4 decimal places. a. What are the correct hypotheses? Null: Ο Ho: μ = 0 O Ho: ₁ = ₂ Ho: M₁ M₂ Ο Ho: μ = 22.3 Alternative: Ο HA: Mı # με Ο HA: μ. > με HA: ₁2 HA: < 0 Ο HA: μ + 22.3 b. What is the correct distribution to use for the test? Select an answer c. What is the test statistic? d. What is the p-value? p-value = e. The correct decision is to Select an answer f. INTERPRET the conclusion in the context of the…
- According to the American Time Use Survey, the typical American spends 183 minutes per day watching television. A college dean wondered if college students were different. He conducted a survey of 50 college students and found they had a mean of 165 minutes per day with a standard deviation of 72. To test this hypothesis, find P-value and rind to the nearest thousandth. Using the a=0.05, write a conclusion for the hypothesis test for question above, state whether or not you are rejecting hypothesis and what the hypothesis is in words, write a sentence that the researcher could share with colleagues without using technical terms for statisticsPlease provide a clear and complete solution. Answer fast for I have 30 minutes left. Thank you very much.Your friend Mona claims that the average student debt immediately after graduation in the United States is $30,000. You want to see if your university has lower student debt at graduation. To test this, you randomly collect data from 169 students who recently graduated. The average of your sample is $29,321, with a sample standard deviation of $6,257. Using this data to perform the hypothesis test with H0: mean=30,000 vs H0: mean<30,000. What is the p-value of this test, and what is the conclusion of this test at the alpha=0.10 level?
- New York State claims that the average cost for gasoline is less than $2.60 per gallon.You think this information is incorrect, so you travel to 41 different gas stations and findthat the mean cost of gas is $2.57 per gallon. Assume the population standard deviationis $0.25 per gallon. Is there enough evidence to support the claim for ? = 0.01? Be sureto identify the null and alternative hypotheses.The College Board reports that the scores on the math SAT were normally distributed with a mean of 525. A random sample of 15 students at your high school had a mean of 550 and a sample standard deviation of 105. What values of the sample mean would lead to rejection of the null hypothesis at the 0.01 level of significance?The school nurse thinks the average height of 10th graders have increased. The average height of a 10th grader five years ago was 145 cm with a standard deviation of 20 cm. She takes a random sample of 200 students and finds that the average height of her sample is 147 cm. Are 10th graders now taller than they were before? Conduct a hypothesis test using a .05 significance level.
- You survey 40 elementary school students on their current reading habits and find that they read 56 minutes per day on average. Assume the standard deviation of reading time per day is 38 minutes. Go through the six steps for testing a hypothesis. Step 1: (Re)write the null and alternative hypotheses. H0 = µ ≤ 48 HA = µ > 48 You choose a = 0.01 as the significance level. Step 2: Calculate the test statistic. Step 3: Find the critical value(s). Step 4: Compare the test statistic to the critical value(s). Step 5: State the conclusion.Historically, the mean age of all Memphis residents has been 40 years. You believe this has changed. You take a random sample of 130 Memphis residents and find that their mean age is 36 years with a standard deviation of 17 years. Can you conclude that the mean age of all Memphis residents has changed from 40 years. For the hypothesis testing scenario above, compute the test statistic. For the hypothesis testing scenario above, you conclude that the mean age of all Memphis residents has changed from 40 years at the 0.05 significance level. Compute the 95% confidence interval to estimate the mean age of all Memphis residents.