2. The Fruit Zoot company claims a packet of Apple Zooties contains an average of 12 fruitsnacks, but a local pediatrician thinks the true average is different. He takes a random sample of15 packets and find an average of 10 fruit snacks with a standard deviation of 1.5. Assume thepopulation is normally distributed with an unknown variance.a) Conduct a hypothesis test for the pediatrician's claim. Use the rejection region approach.b) Construct a 95% confidence interval for the true mean number of fruit snacks. Compare thisinterval to your conclusion to the test in part a.

Answered: Image /qna-images/answer/a71960f4-164d-4092-9aac-55648faabf13.jpg

Answered: 2. The Fruit Zoot company claims a…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A custodian wishes to compare two competing floor waxes to decide which one is best. He believes that the mean of WaxWin is less than the mean of WaxCo. In a random sample of 18 floors of WaxWin and 10 of WaxCo. WaxWin had a mean lifetime of 22.1 with a standard deviation of 5.6 and WaxCo had a mean lifetime of 28.5 with a standard deviation of 7.1. Perform a hypothesis test using a significance level of 0.05 to help him decide. Let WaxWin be sample 1 and WaxCo be sample 2 The correct hypotheses are: O Ho: H1 µ2(claim) O Ho: µ1 2 H2 HA: H1 < µ2(claim) O Ho: µ1 = µ2 HA: H1 7 M2(claim) Since the level of significance is 0.05 the critical value is -1.751 The test statistic is: (round to 3 places) The p-value is: (round to 3 places) The decision can be made to: O reject Ho O do not reject Ho The final conclusion is that: O There is enough evidence to reject the claim that the mean of WaxWin is less than the mean of WaxCo. O There is not enough evidence to reject the claim that the mean of…
Ms. Rains believes more seniors from PLD plan to attend college than seniors at Lafayette. In a random sample of 35 PLD seniors, 21 responded they plan to attend college. In a random sample of 32 Lafayette seniors only 17 responded they plan to attend college. Using a 95% confidence interval, predict the true difference in the proportion of seniors at PLD and Lafayette who plan to attend college and comment on Ms. Rains' belief.
The principal at a large high school claims that students spend at least 10 hours per week doing homework, on average. To investigate this claim, an AP Statistics class selected a random sample of 250 students from their school and asked them how long they spent doing homework during the last week. The sample mean was 10.2 hours and the sample standard deviation was 4.2 hours. Construct and interpret a 95% confidence interval for the mean time that students at this school spent doing homework in the last week. Based on your interval, what can you conclude about the principal's claim? What is your t* value? Remember to use the conservative DF value.
In an engineering class, students want to determine the difference in the mean strength between two different types of building materials—Type A, which is relatively inexpensive, and Type B, which is slightly more expensive. The students take a random sample of 60 of the Type A building material and a random sample of 27 of the Type B building material. Given the mean strength for each sample, the students construct a 95% confidence interval for the difference in population means. (a) Determine the degrees of freedom for the conservative approach for finding the t critical value. (b) Find the t critical value associated with 95% confidence. (Use a table or SALT. Round your answer to two decimal places.)
The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maximum possible rating is 10. Suppose a simple random sample of 50 business travelers is selected and each traveler is asked to provide a rating for the Miami International Airport. The ratings obtained from the sample of 50 business travelers provides a mean of 6.34 and a standard deviation of 2.163. Develop a 95% confidence interval estimate of the population mean rating for Miami.
A single burger contains a mean of 100 g of cooked beef with a standard deviation of 5 g. The restaurant owner claims that new triple burger contains three times as much beef. The table shows a sample of masses from 30 triple burgers. Does the sample support the claim at a 95% confidence level? Triple Burger Beef Mass (g) 292 297 295 290 290 294 291 296 282 291 289 291 290 292 295 289 289 294 298 292 294 295 285 297 296 287 302 289 290 290
An economist is studying salaries for high technology companies and wants to test the claim that the average salary for high tech employees is less than $70,000. The economist selects a sample of 35 random employees from various high tech companies and records their salaries. Based on past studies, the economist determines that the population standard deviation is $8,500. The economist conducts a one-mean hypothesis test at the 5% significance level to test the claim that the average salary for high tech employees is less than $70,000. Which is the correct setup for the null and alternative hypothesis for this example? Select the correct answer below: H0:μ=70,000; Ha:μ<70,000, which is a left-tailed test. H0:μ=70,000; Ha:μ>70,000, which is a right-tailed test H0:μ<70,000; Ha:μ=70,000, which is a left-tailed test. H0:μ>70,000; Ha:μ=70,000, which is a left-tailed test.
A sports news station wanted to know whether people who live in the North or the South are bigger sports fans. For its study, 121 randomly selected Southerners were surveyed and found to watch a mean of 4.6 hours of sports per week. In the North, 176 randomly selected people were surveyed and found to watch a mean of 3.2 hours of sports per week. Find a 90% confidence interval for the true difference between the mean numbers of hours of sports watched per week for the two regions if the South has a population standard deviation of 1.6 hours per week and the North has a population standard deviation of 1.4 hours per week. Let Population 1 be people who live in the South and Population 2 be people who live in the North. Round the endpoints of the interval to one decimal place, if necessary. Lower endpoint: Upper endpoint:
A custodian wishes to compare two competing floor waxes to decide which one is best. He believes that the mean of WaxWin is less than the mean of WaxCo. In a random sample of 20 floors of WaxWin and 17 of WaxCo. WaxWin had a mean lifetime of 26.4 with a standard deviation of 8.6 and WaxCo had a mean lifetime of 27.2 with a standard deviation of 6.6. Perform a hypothesis test using a significance level of 0.05 to help him decide. Let WaxWin be sample 1 and WaxCo be sample 2 The correct hypotheses are: Ο Ηο: μι < με HA₁₂(claim) Ο Ηρ: με 2 με HA: 1₂(claim) O Ho: ₁ = ₂ HAM12(claim) Since the level of significance is 0.05 the critical value is -1.69 (round to 3 places) The test statistic is: The p-value is: (round to 3 places)
A technician compares repair costs for two types of microwave ovens (type I and type II). He believes that the repair cost for type I ovens is greater than the repair cost for type II ovens. A sample of 35 type I ovens has a mean repair cost of $80.39. The population standard deviation for the repair of type I ovens is known to be $24.63. A sample of 31 type II ovens has a mean repair cost of $73.47. The population standard deviation for the repair of type II ovens is known to be $10.42. Conduct a hypothesis test of the technician's claim at the 0.05 level of significance. Let i be the true mean repair cost for type I ovens and µz be the true mean repair cost for type II ovens. Step 1 of 5: State the null and alternative hypotheses for the test.
A sporting equipment company manufactures 15 pond dumbbells .To ensure that the production line is functioning properly , 50 dumbbells were selected at random , and their weight were recorded .The observed sample mean weight of the 50 dumbbells was 16.87 lb and the observed sample standard deviation was 1.44 lb. A Construct a 90% confidence interval for the mean of the distribution of the weight of the manufactured dumbbells
• Example: A sample of 25 physically and mentally healthy males participated in a sleep experiment in which the percentage of each participant's total sleeping time spent in a certain stage of sleep was recorded. • The variance computed from the sample data was 2.25. Construct 95 percent confidence intervals for o².

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