A simple random sample of 56 students who picked up all tests they completed was selected, and the mean score on final exam for this sample of students was 83 with astandard deviation of 10.4. An independent simple random sample of 51 students who had one or more tests that were not picked up was selected, and the mean scoreon the final exam for this sample of students was 67 with a standard deviation of 24.2. Both distributions are skewed heavily to the left [note: this is the same informationprovided in question above]. If appropriate, use this information to calculate and interpret a 99% confidence interval for the difference in the mean score on the finalexam for all students who picked up all their tests and the mean score on the final exam for all students who had one or more tests that were not picked up.17Yes, because both samples are larger than 40 and there are 2 simple random samplesYes, because any sample size will work and there are 2 simple random samplesYes, because only sample sizes of 15 are needed and there are 2 simple random samples,What are the confidence interval values using 99% confidence?(11.615, 26.385)(6.2876, 25.712)(13.506, 20.494)O(6.568, 25.434)(6.6985, 25.302)-D

Confidence interval is used to determine the parameter range where it lies. It explains true mean…

Answered: A simple random sample of 56 students…

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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Suppose that El Camino is studying the average number of hours students study for classes per week. According to research from 2010 El Camino students spent an average of 21 hours a week studying for classes. El Camino wants to study if this average has changed. To do this they take a random sample of 81 current students and find that the average student spent 19.3 hours studying with a standard deviation of 6.4 hours. (Note: You may assume there are approximately 20,000 El Camino students). (a) Carry out a hypothesis test at α = .05 to test the claim that the average number of hours per week that El Camino students study has changed since 2010. (b) Describe what it would mean in this situation to make a Type I error (c) Describe what it would mean in this situation to make a Type II error (d) Construct a 95% confidence interval for the average number of hours studied by El Camino students per week. Make sure to interpret your interval.
A professor believes the class he is teaching in the current semester is below the average of the classes he has taught. In order to test this hypothesis, he analyses the scores of midterm of all 100 students he teaches this semester. Their average score of this exam is 79 out of 100, with a standard deviation of 12. The mean score of the midterm for all students that he has taught for the last thirty-two years is 83 out of 100, with a standard deviation of 14. Use an alpha level of 0.01 for this analysis. Part B.State the null hypothesis and the alternative hypothesis:
A pizza delivery chain advertises that it will deliver your pizza in 20 minutes from when the order is placed. Being a skeptic, you decide to test and see if the mean delivery time is actually more than 20minutes. For the simple random sample of 6customers who record the amount of time it takes for each of their pizzas to be delivered, the mean is 21.9 minutes with a standard deviation of 2.1minutes. Assume that the population distribution is approximately normal. Perform a hypothesis test using a 0.025level of significance. Step 3 of 3 : Draw a conclusion and interpret the decision. Answer
Fran is training for her first marathon, and she wants to know if there is a significant difference between the mean number of miles run each week by group runners and individual runners who are training for marathons. She interviews 42 randomly selected people who train in groups and finds that they run a mean of 47.1 miles per week. Assume that the population standard deviation for group runners is known to be 4.4 miles per week. She also interviews a random sample of 47 people who train on their own and finds that they run a mean of 48.5 miles per week. Assume that the population standard deviation for people who run by themselves is 1.8 miles per week. Test the claim at the 0.01 level of significance. Let group runners training for marathons be Population 1 and let individual runners training for marathons be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.
A half-century ago, the mean height of women in a particular country in their 20s was 62.4 inches. Assume that the heights of today's women in their 20s are approximately normally distributed with a standard deviation of 1.67 inches. If the mean height today is the same as that of a half-century ago, what percentage of all samples of 27 of today's women in their 20s have mean heights of at least 63.02 inches? About% of all samples have mean heights of at least 63.02 inches. (Round to one decimal place as needed.)
A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 10 days following no advertisements, the mean was 18.3 purchasing customers with a standard deviation of 1.8 customers. On 7 days following advertising, the mean was 19.4 purchasing customers with a standard deviation of 1.6 customers. Test the claim, at the 0.02 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2. Step 3 of 3: Draw a…
A car company claims that its new SUV gets better gas mileage than its competitor's SUV. A random sample of 35 of its SUVs has a mean gas mileage of 16.7 miles per gallon (mpg). The population standard deviation is known to be 0.9 mpg. A random sample of 42 competitor's SUVS has a mean gas mileage of 16.4 mpg. The population standard deviation for the competitor is known to be 0.6 mpg. Test the company's claim at the 0.02 level of significance. Let the car company's SUVs be Population 1 and let the competitor's SUVS be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Họ :-1 = M2 Ha М2
Pilots who cannot maintain regular sleep hours due to their work schedule often suffer from insomnia. A recent study on sleeping patterns of pilots focused on quantifying deviations from regular sleep hours. A random sample of 28 commercial airline pilots was interviewed, and the pilots in the sample reported the time at which they went to sleep on their most recent working day. The study gave the sample mean and standard deviation of the times reported by pilots, with these times measured in hours after midnight. (Thus, if the pilot reported going to sleep at p.m., the measurement was -1.) The sample mean was 0.8 hours, and the standard deviation was 1.6 hours. Assume that the sample is drawn from a normally distributed population. Find a 90% confidence interval for the population standard deviation, that is, the standard deviation of the time (hours after midnight) at which pilots go to sleep on their work days. Then give its lower limit and upper limit.
A nurse is interested in the amount of time patients spend exercising per day. According to a recent study, the daily workout time per adult follows an approximately normal distribution with a mean of 94 minutes and a standard deviation of 27minutes. If the nurse randomly samples patients in her office to analyze their exercise time and gets a standard error of 3minutes, how many patients did she sample?
In a school district, all sixth grade students take the same standardized test. The superintendant of the school district takes a random sample of 3030 scores from all of the students who took the test. She sees that the mean score is 169169 with a standard deviation of 7.06747.0674. The superintendant wants to know if the standard deviation has changed this year. Previously, the population standard deviation was 1313. Is there evidence that the standard deviation of test scores has decreased at the α=0.005α=0.005 level? Assume the population is normally distributed. Step 1 of 5: State the null and alternative hypotheses. Round to four decimal places when necessary. Step 2 of 5: 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 of 5: Determine the value of the test statistic. Round your answer to three decimal places. Step 4 of 5: Make the…
The salaries of professional baseball players are heavily skewed right with a mean of $3.2 million and a standard deviation of $2 million. The salaries of professional football players are also heavily skewed right with a mean of $1.9 million and a standard deviation of $1.5 million. A random sample of 40 baseball players' salaries and 35 football players' salaries is selected. The mean salary is determined for both samples. Let -, represent the difference in the mean salaries for baseball and football players. Which of the following represents the shape of the sampling distribution for ,-7,? skewed right since the populations are both right skewed skewed right since the differences in salaries cannot be negative approximately Normal since both sample sizes are greater than 30 approximately Normal since the sum of the sample sizes is greater than 30
A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 13 days following no advertisements, the mean was 23.9 purchasing customers with a standard deviation of 1.9 customers. On 6 days following advertising, the mean was 24.7 purchasing customers with a standard deviation of 1.6 customers. Test the claim, at the 0.01 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2. Step 3 of 3: Draw a…

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