A random sample of 100 workers with children in day care shows a mean day-care cost of Rs. 2600 and a
standard deviation of Rs.500. Verify the department’s claim that the mean exceeds Rs. 2500, at the 0.05
level with this information.
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3, Hypothesis testing for the mean (gis known) Find the P-value for a two-tailed hypothesis test with a standardized test statistic of z 1.64. Decide whether to reject Ho when the level of significance is α a. 0.10. b. Find the P-value for a right-tailed hypothesis test with a standardized test statistic of z 1.64. Decide whether to reject Ho when the level of significance is a0.10. Homeowners claim that the mean speed of automobiles traveling on their street is...
Conduct a hypothesis test for each problem, using the traditional method. Show the 5 steps and all work for each hypothesis test. Be sure you select the correct test to use for each problem. 1. A telephone company claims that less than 30% of all college students have a limited number of text messages per month. A random sample of 150 students revealed that 41 of them have a limited number. Test the company's claim at the 0.01 level of...
Suppose the null hypothesis is Ho : µ = 500 against Ha : > µ = 500 , and the significance level for this testing is 0.05. The population in question is normally distributed with standard deviation 100. A random sample of size n=25 will be used. If the true alternative mean is 550, then the probability of committing the type II error is ____.
Hypothesis Testing Worksheet 1. The cold medicine Dozenol list 600 milligrams of acetaminophen per fluid Ounce as an active ingredient. The Food and Drug Administration tests 65 one-ounce samples of the medicine and finds that the mean amount of acetaminophen for the sample is 589 milligrams with a standard deviation of 21 milligrams. Test the claim of the FDA that the medicine does not contain the required amount of acetaminophen at the level of significance 0.05. 2. The Medassist Pharmaceutical...
PLEASE INCLUDE GRAPH!!! and please answer correctly 3 2004 Hypothesis test for 1 mean Page 406-413 A fast food restaurant claims that the mean waiting time is greater than random sample of 50 customers revealed a sample mean waiting time of 2.0 the mean waiting time is greater than 3.2 minutes (original claim). A deviation of 0.486 minutes. revealed a sample mean waiting time of 2.8 minutes with a sample standard Test the restaurants claim at a a= 0.05 level...
17) A store manager claims that the standard deviation of the number of customers per day is more than 8. A random sample of 51 days has a mean number of customers 498 with a standard deviation 8.2. Test the claim using a 0.05 level of significance.
Garrison Keller claims the children of Lake Wobegon are above average. You take a simple random sample of 9 children from Lake Wobegon and measure their intelligence with a Wechsler test and find the following scores: {116, 128, 125, 119, 89, 99, 105, 116, and 118}. The mean of this sample (? ̅) is 112.8. We know Wechsler scores are scaled to be normally distributed with a mean of 100 and standard deviation of 15. Is this sample mean sufficiently...
7) A pizza chain advertises that they will deliver a pizza in no more than 20 minutes from when an order is placed. However, for a random sample of 7 orders, the mean delivery time was 22.7 minutes with a standard deviation of 4.3 minutes. At the 0.05 significance level, test the claim that the mean pizza delivery time exceeds 20 minutes.
3. Test the claim that a population mean exceeds 40. You have a sample of 40 items for which the sample mean x 42 and sample standard deviation s- 7. Use significance level α 0.05.
A psychologist obtains a random sample of 20 mothers in the first trimester of their pregnancy. The mothers are asked to play Mozart in the house at least 30 minutes each day until they give birth. After 5 years, the child is administered an IQ test. It is known that IQs are normally distributed with a mean of 100. If the IQs of the 20 children in the study result in a sample mean of 104.2 and sample standard deviation...