A computer company stated that children uses their products 6.77 hours per week. You gathered 30 independent samples from a group of children using their products and obtained an average 7.8 hours per week, with a sample standard deviation of 2.9 hours and at alpha = 0.10, compute the critical value
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A computer company stated that children uses their products 6.77 hours per week. You gathered 30 independent samples from a group of children using their products and obtained an average 7.8 hours per week, with a sample standard deviation of 2.9 hours and at alpha = 0.10, compute the critical value
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- You are performing a right-tailed matched-pairs test with 9 pairs of data.If α=.005α=.005, find the critical value, to two decimal places.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 40 drivers over the age of 60 and finds that the mean number of miles driven is 22.7. The population standard deviation is 2.9 miles. At =α0.01, 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 critical value method with tables. State the hypotheses and identify the claim with the correct hypothesis.A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for SUVs equipped with the tires. The mean braking distance for SUVs equipped with tires made with compound 1 is 65 feet, with a population standard deviation of 13.6. The mean braking distance for SUVS equipped with tires made with compound 2 is 69 feet, with a population standard deviation of 8.5. Suppose that a sample of 55 braking tests are performed for each compound. Using these results, test the claim that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. Let μ₁ be the true mean braking distance corresponding to compound 1 and μ₂ be the true mean braking distance corresponding to compound 2. Use the 0.05 level of significance. Step 3 of 5 Find the p-value associated with the test statistic. Round your answer to four decimal places. Answer How to enter your answer (opens in…
- Currently patrons at the library speak at an average of 61 decibels. Will this average increase after the installation of a new computer plug in station? After the plug in station was built, the librarian randomly recorded 57 people speaking at the library. Their average decibel level was 63.8 and their standard deviation was 9. What can be concluded at the the αα = 0.10 level of significance?A nationwide study of undergraduate students reported that the mean number of drinks consumed per week during the spring semester is 7.96. The mean number of drinks consumed per week at USC is 7.64 (s.d.=2.55, N=412 Health services is concerned that USC students are consuming significantly more alcohol per week than the national average. Using an alpha level of .05, Is there sufficient evidence to be concerned? Be sure to select the correct critical value for the alternative hypothesis, and then use this evidence to make your conclusionA company that manufactures light bulbs claims that its light bulbs last an average of 1,150 hours. A sample of 25 light bulbs manufactured by this company gave a mean life of 1067 hours and a standard deviation of 137 hours. A consumer group wants to test the hypothesis that the mean life of light bulbs produced by this company is less than 1,150 hours. The significance level is 5%. Assume the population is normally distributed. What is the critical value of t? O-2.787 O-1.711 O-1.704 -2.797 What is the value of the test statistic, t, rounded to three decimal places? What is the p-value for this hypothesis test, rounded to four decimal places? Should you reject or fail to reject the null hypothesis in this test? Does the data provide evidence to contradict the company's claim about the average lifetime of their light bulbs?
- A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for SUVs equipped with the tires. The mean braking distance for SUVS equipped with tires made with compound 1 is 68 feet, with a population standard deviation of 13.9. The mean braking distance for SUVs equipped with tires made with compound 2 is 73 feet, with a population standard deviation of 13.7. Suppose that a sample of 72 braking tests are performed for each compound. Using these results, test the claim that the braking distance for SUVS equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. Let μ, be the true mean braking distance corresponding to compound 1 and μ₂ be the true mean braking distance corresponding to compound 2. Use the 0.05 level of significance. Step 1 of 5: State the null and alternative hypotheses for the test.The braking ability was compared for two car models. Random samples of two cars were selected. The first random sample of size 64 cars yield a mean of 36 and a standard deviation of 8. The second sample of size 64 yield a sample mean of 33 and a standard deviation of 8. Do the data provide sufficient evidence to indicate a difference between the mean stopping distances for the two models? Use Alpha= 0.01. Ho: µ1 – µ2 = 0 vs. Ha: µ1 – µ2 + 0 .p-value = 0.0017. Reject Ho Но: Д, — м2 — 0 vs. Ha:M1 — M2 + 0. p-value — 0.034. Ассеpt Ho O Ho:µ1 – Hz Но: И — 2 — 0 vs. Ha:M, — нz + 0. p-value —D 0.0017. Ассept Ho Ho: H1 – U2 = 0 vs. Ha: µ1 – µ2 + 0. p-value = 0.034. Reject HoOne year consumers spent an average of $22 on a meal at a resturant. Assume that the amount spent on a resturant meal is normally distributed and that the standard deviation is $3.Complete parts (a) through (c) below. a. What is the probability that a randomly selected person spent more than $24?
- The herbal supplement ginkgo biloba is advertised as producing an increase in physical strength and stamina. To test this claim, a sample of n = 120 adults is obtained and each person is instructed to take the regular daily dose of the herb for a period of 30 days. At the end of the 30 days each person is tested on a standard treadmill task for which the average age-adjusted score is μ = 15. The individuals in the sample produce a mean score of M = 17.4 with SS = 1260. Are these data sufficient to conclude that the herb had a statistically significant effect? Conduct as a two-tailed test, using all hypothesis steps with p < .05. BONUS POINTS: Calculate the 95% confidence intervalA researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for SUVs equipped with the tires. The mean braking distance for SUVs equipped with tires made with compound 1 is 65 feet, with a population standard deviation of 13.6. The mean braking distance for SUVs equipped with tires made with compound 2 is 69 feet, with a population standard deviation of 8.5. Suppose that a sample of 55 braking tests are performed for each compound. Using these results, test the claim that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. Let μ₁ be the true mean braking distance corresponding to compound 1 and μ₂ be the true mean braking distance corresponding to compound 2. Use the 0.05 level of significance. Step 5 of 5: State the conclusion of the hypothesis test. Answer Tables Keypad Keyboard Shortcuts Previous Step Answers There is sufficient evidence…Suppose there are two different vaccines for Covid, Vaccine X and Vaccine Y. An interesting question is which vaccine has a higher 6-month antibody effectiveness quotient (6AEQ). To examine this we randomly select 78 recipients of vaccine X and 93 recipients on vaccine Y. The vaccine X recipients had a mean 6AEQ of x = 151. The vaccine Y recipients had a mean 6AEQ of y = 148. It is recognized that the true standard deviation of 6AEQ for vaccine X recipients is 0x = 8.7 while it is recognized that the true standard deviation of 6AEQ for vaccine Y recipients is dy = 11.5. The true (unknown) mean 6AEQ for vaccine X recipients is μx, while the true (unknown) mean 6AEQ for vaccine Y recipients is y. 6AEQ measurements are known to be a normally distributed. In summary: Type Sample Size Sample Mean Standard Deviation Vaccine X 78 Vaccine Y 93 151 148 8.7 11.5 a) Calculate the variance of the random variable X which is the mean of the 6AEQ measurements of the 78 vaccine X recipients.…