The null hypothesis asserts that the variance is less than 6. A random sample size of 26 is drawn from the population and yieldds a sample mean of 12.95 and a standard deviation of 5.5. What is the test statistic when alpha = 0.05?
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The null hypothesis asserts that the variance is less than 6. A random sample size of 26 is drawn from the population and yieldds a sample mean of 12.95 and a standard deviation of 5.5. What is the test statistic when alpha = 0.05?
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- A psychologist wants to test whether there is any difference in puzzle-solving abilities between boys and girls. Independent samples of 10 boys and 14 girls were chosen at random. The boys took a mean of 30 minutes to solve a certain puzzle with a standard deviation of 8 minutes. The girls took a mean of 34 minutes to solve the same puzzle with a standard deviation of 5 minutes. Assume that the two populations of completion times are normally distributed and that the population variances are equal. Construct a 95% confidence interval for the difference µ, - µ, between the mean puzzle-solving times for boys (u,) and for girls (H2). Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your responses to at least two decimal places. (If necessary, consult a list of formulas.) What is the lower limit of the 95% confidence interval? What is the upper limit of the 95% confidence interval?A company is doing a hypothesis test on the variation in quality from two suppliers. Both distributions are normal, and the populations are independent. Use a = 0.01. A sample of 26 products were selected from Supplier 1 and a standard deviation of quality was found to be 6.0042. A sample of 25 products were selected from Supplier 2 and a standard deviation of quality was found to be 1.802. Test to see if the variance in quality for Supplier 1 is larger than Supplier 2. What are the correct hypotheses? Note this may view better in full screen mode. Select the correct symbols in the order they appear in the problem. Ho: Select an answer ✓ ? Select an answer H₁: Select an answer ? ✓ Select an answer Based on the hypotheses, compute the following: Round answers to at least 4 decimal places. The test statistic is = The p-value is The decision is to Select an answer The correct summary would be: Select an answer quality for Supplier 1 is larger than Supplier 2. ✓that the variance inColonial Funds claims to have a bond fund which has performed consistently throughout the past year. The variance of the share price is claimed to be 0.17 To test this claim, an investor randomly selects 24 days during the last year to check the performance of the fund. He finds an average share price of $15.80 with a standard deviation of 0.2113. Can the investor conclude that the variance of the share price of the bond fund is different than claimed at α=0.05 Assume the population is normally distributed. Step 1: State the null and alternative hypotheses. Round to four decimal places when necessary. Step 2: Determine the critical value(s) of the test statistic. If the test is two-tailed, separate the values with a comma. Round your answer to three decimal places. Step 3: Determine the value of the test statistic. Round your answer to three decimal places. Step 4: Make the decision: Reject Null Hypothesis or Fail to Reject Null Hypothesis Step 5: What is the conclusion? 1)There is…
- X is a normal variable with mean of 8 and variance of 97. What is the z-score when x = 3?The concentration of benzene was measured in units of milligrams per liter for a simple random sample of five specimens of untreated wastewater produced at a gas field. The sample mean was 4.6 with a sample standard deviation of 2.7. Seven specimens of treated wastewater had an average benzene concentration of 7.9 with a standard deviation of 2.3. It is reasonable to assume that both samples come from populations that are approximately normal. Can you conclude that the mean benzene concentration differs between treated water and untreated water? Let h1 denote the mean benzene concentration for untreated water and h2 denote the mean benzene concentration for treated water. Use the a=0.05 level the p-value method with the TI-84 Plus calculator. please show p value!Do men take a different amount of time than women to get out of bed in the morning? The 45 men observed averaged 6.1 minutes to get out of bed after the alarm rang. Their standard deviation was 5.6. The 42 women observed averaged 6 minutes, and their standard deviation was 4 minutes. What can be concluded at the a = 0.01 level of significance? Assuming Equal Variances a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho: Select an answer Select an answer Select an answer v (please enter a decimal) H1: Select an answer v Select an answer Select an answer v (Please enter a decimal) c. The test statistic ? d. The test statistics (please show your answer to 3 decimal places.) N
- Previous research has shown that the mean number of ballpoint pens Americans use per year is 4.3 with a standard deviation of 1.17. You believe that amount is too high since computers are used so often this day and age. You randomly select 20 people and find the average to be 3.8 ballpoint pens per year. Assuming normality, is there evidence to show that the mean has decreased? What is the decision for this hypothesis test? Previous research has shown that the mean number of ballpoint pens Americans use per year is 4.3 with a standard deviation of 1.17. You believe that amount is too high since computers are used so often this day and age. You randomly select 20 people and find the average to be 3.8 ballpoint pens per year. Assuming normality, is there evidence to show that the mean has decreased? What is the decision for this hypothesis test? Reject the null because the test statistic is in the critical region and the p-value is less than alpha. Reject the null because the test…Records indicate that the mean weight of mature rainbow trout in Eagle Creek is 1.75 kg with a standard deviation of 0.37 kg. After years of marked oxygen depletion from pollutants in the creek, you want to perform a hypothesis test to see if the standard deviation, ơ, of weights has changed. To do so, you measure the weights of 19 randomly chosen mature rainbow trout from the creek and find that the sample standard deviation is 0.49 kg. Under the assumption that current weights of mature rainbow trout in the creek follow a normal distribution, you will perform a chi-square test. Find x, the value of the test statistic for your chi-square test. Round your answer to three or more decimal places. 2You want to examine the relationship between dropping out of high school and delinquency. You have a random sample of 11 students. You have the number of delinquent offenses that each student reported committing in the year before and after dropping out of school. The following table shows the data. Student Before dropping out After dropping out 1 5 7 2 9 5 3 2 3 4 7 7 5 8 11 6 11 13 7 8 4 8 8 10 9 5 7 10 2 1 11 9 3 From performing a hypothesis test to determine if the mean number of delinquent differs after dropping out, what is the obtained t?
- A random sample of 10 subjects have weights with a standard deviation of 10.5816 kg. What is the variance of their weights? Be sure to include the appropriate units with the result.Daily Driving The average number of miles a person drives per day is 24. A researcher wishes to see if people over age 60 drive less than 24 miles per day. She selects a random sample of 32 drivers over the age of 60 and finds that the mean number of miles driven is 22.9. The population standard deviation is 3.1 miles. At α=0.05, is there sufficient evidence that those drivers over 60 years old drive less than 24 miles per day on average? Assume that the variable is normally distributed. Use the P-value method with a graphing calculator. Find the P-value. Round it to four decimal places.pounds, with a standard deviation of 15.81 pounds. At the 0.05 level of significance, test the claim that the weights of football players have a greater variance than the weights of baseball deviation of 40.09 pounds. A random sample of 25 baseball players had a mean weight of 202.9 6. A random sample of 48 football players had a mean weight of 246.6 pounds, with a standard ith a players. ie th sufficient evidence to conclude that Step 5: There