**Text Transcription and Explanation for an Educational Website**---**Hypothesis Testing and Analysis of Weight-Loss Program**An advertisement for a new weight-loss program claims that participants lose an average of 18 pounds during their first month in the program. A skeptical medical professional believes that this figure is too high and wishes to conduct a hypothesis test at the $ \alpha = 0.01 $ level of significance to determine if they are correct. Suppose it is known that the population standard deviation for weight loss is 2.7 pounds. (Round your results to four decimal places.)**Which would be the correct hypotheses for this test?**- $ H_0: \mu = 18, \, H_1: \mu > 18 $- $ H_0: \mu < 18, \, H_1: \mu = 18 $- $ H_0: \mu \neq 18, \, H_1: \mu = 18 $- $ H_0: \mu = 18, \, H_1: \mu \neq 18 $- $ H_0: \mu = 18, \, H_1: \mu < 18 $**A random sample of 54 participants lost an average of 17 pounds. Find the test statistic:**[Input Box for Test Statistic]**Give the P-value:**[Input Box for P-value]**Which is the correct result?**- [ ] Reject the Null Hypothesis- [ ] Do not Reject the Null Hypothesis**Which would be the appropriate conclusion?**- [ ] There is significant evidence to suggest that the actual mean weight loss for participants is less than 18- [ ] There is not significant evidence to suggest that the actual mean weight loss for participants is less than 18---**Note for Educators:**This problem is designed to help students practice hypothesis testing. The scenario presented involves determining whether the average weight loss claimed by a program is accurate or inflated. Students must identify the correct hypotheses, calculate the test statistic, determine the P-value, and draw a conclusion based on the significance level provided. This exercise reinforces understanding of hypothesis testing, including formulating null and alternative hypotheses, using sample data, and interpreting statistical outcomes in the context of the significance level.

Given that:Hypothesized mean is .The level of significance is .The population standard deviation is…

Answered: An ad for a new weight-loss program…

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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When engaging in weight-control (fitness/fat burning) types of exercise, a person is expected to attain approximately 60% of his or her maximum heart rate. For 20-year-olds, this rate is approximately 120 bpm. A simple random sample of thirty 20-year-olds was taken, and the sample mean was found to be 107 bpm, with a standard deviation of 45 bpm. Researchers wonder if this is evidence to conclude that the expected level is actually lower than 120 bpm. Report a 95% confidence region for the mean level of heart rate for this group. the mean level of heart rate is lower than 120.96 the mean level of heart rate is greater than 114.47 the mean level of heart rate is greater than 117.67 the mean level of heart rate is lower than 114.47
Under what circumstances is a t statistic used instead of a z-score for a hypothesis test? Justin wants to know whether a commonly prescribed drug does improve the attention span of students with attention deficit disorder (ADD). He knows that the mean attention span for students with ADD who are not taking the drug is 2.3 minutes long. His sample of 12 students taking the drug yielded a mean of 4.6 minutes. Justin can find no information regarding σx , so he calculated s2x =1.96. Determine the critical region using a one-tailed test with alpha = .05. Conduct the hypothesis test (Do the math and compare the t-critical and t-obtained values). State your conclusions in terms of H0 (Should you reject the H0 or fail to reject/accept the H0). Based on your analysis, is there a relationship between the drug and attention span?
An article in the San Jose Mercury News stated that students in the California state university system take an average of 4.5 years to finish their undergraduate degrees. Suppose you believe that the average time is longer. You conduct a survey of 42 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level? Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) + Part (a) O Part (b) O Part (c) + Part (d) O Part (e) O Part (f) O Part (g) O Part (h) O Part (i) Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your lower and upper bounds to two decimal places.) 95% C.I. 2.0195 4.7261 5.4739 Additional Materials
The mean annual tuition and fees for a sample of 14 private colleges in California was $37,900 witrh a standard deviation of $7,200. A dotplot shows that it is reasonable to assume that the population is approxiamately normal. Can you conclude that the mean tuition and fees for private institutions in California differs from $35,000? a. State the null and alternate hypothesis. b. Compute the value of the t statistic. How many degrees of fredom are there?
Suppose you are interested in testing whether average monthly gasoline expenditure differs from $50 at the 5% significance level. You take a sample of 30 and calculate a sample mean of 51 and sample standard deviation of 12. Given this information the null and alternative hypotheses are as follows: H0: μ=50 H1: μ≠50 What is the p-value for this test? (Round to four decimal places. )
A survey of nonprofit organizations showed that online fundraising increased in the past year. Based on a random sample of 53 nonprofit organizations, the mean one-time gift donation in the past year was $80,with a standard deviation of $13. . If you test the null hypothesis at the 0.05 level of significance, is there evidence that the mean one-time gift donation is greater than $75? a. State the null and alternative hypotheses. Choose the correct answer below. b. Find the test statistic. c. Find the p-value for this hypothesis test.
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It's proposed that the average life expectancy these days is 81.6, but recent data suggests it might be different from that. A sample size of 89 countries found a sample mean life expectancy of 79.2 years with a standard deviation of 18.6 years. You want to formulate and test a hypothesis that can be used to challenge the validity of the claim at a significance level of 0.01. With hypotheses Ho: µ = 81.6 vs H1: µ # 81.6, what is/are the critical value(s)? (Report answer accurate to 2 decimal places.) What is the test statistic for this sample? What is the p-value for this sample? Our decision is to reject the null accept the null fail to reject the null As such, the final conclusion is that there is sufficient evidence to conclude that the population mean life expectancy is equal to 81.6. there is not sufficient evidence to conclude that the population mean life expectancy is equal to 81.6. there is sufficient evidence to conclude that the population mean life expectancy is not equal…
Medicare would like to test the hypothesis that the average monthly rate for one- bedroom assisted-living facility is equal to $3,300. A random sample of 12 assisted- living facilities had an average rate of $3,690 per month. The standard deviation for this sample was $530. Medicare would like to set a = 0.05. Which one of the following statements is true? a. Because the p-value is less than alpha, we fail to reject the null hypothesis and conclude that the average monthly rate for an assisted- living facility is not equal to $3,300. b. Because the p-value is less than alpha, we reject the null hypothesis and conclude that the average monthly rate for an assisted-living facility is not equal to $3,300. O c. Because the p-value is greater than alpha, we fail to reject the null hypothesis and conclude that the average monthly rate for an assisted- living facility is equal to $3,300.
true or false? Suppose a sample size of 87 was used to conduct a right tailed hypothesis test for a single population mean. The null hypothesis was H0: μμ = 45 and the sample mean was 50 and the p-value was 0.03. Then there is a 3% chance that 87 randomly selected individuals from a population with a mean of 45 and the same standard deviation would have a sample mean greater than 50.

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