A brochure inviting subscriptions for a new diet program states that the participants are expected to lose over 22 pounds in five weeks. Suppose that, from the data of the five-week weight losses of 56 participants, the sample mean and sample standard deviation are found to be 23.5 and 10.2, respectively. Could the statement in the brochure be substantiated on the basis of these findings? Test at α = 0.05.
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- A brochure inviting subscriptions for a new diet program states that the participants are expected to lose over 22 pounds in five weeks. Suppose that, from the data of the five-week weight losses of 56 participants, the sample mean and sample standard deviation are found to be 23.5 and 10.2, respectively. Could the statement in the brochure be substantiated on the basis of these findings? Test at α = 0.05.
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- In the year 2032, Katie Ruff is a leading traveling nurse. Katie is interested in reducing the mean recovery time for patients after experiencing a serious injury. Suppose the mean recovery time is presently 8.4 months. Katie takes a random sample of 45 patients that have experienced serious injury to participate in a new treatment program and finds the sample mean is 7.9 months and a sample standard deviation of 1.2 months. Using α = 0.05, answer the following questions. a) What is the setup for your null and alternative hypothesis? b) What is the value of the test statistic? c) Using the appropriate level of confidence, what is the confidence interval? d) What is the interpretation (not your conclusions) in the context of the problem of the confidence interval you obtained in part c? e) What is the p value? f) What is the interpretation (not your conclusions) of the p value in the context of the problem you found in part…At age 9 the average weight for both boys and girls are exactly the same. A random sample of 9 year olds gave the following results.... Boys: Girls:Sample Size: 65 50Mean weight (lbs): 130 125Pop. Variance: 120 130 At alpha 0.05, do the data support the claim that there is a difference in heights?Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 4.8 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 242samples is 5.2 ppm with a variance of 1.001.00. Does the data support the claim at the 0.05 level? Assume the population distribution is approximately normal. Find the value of the test statistic. Round your answer to three decimal places. Specify if the test is one-tailed or two-tailed. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places. Make the decision to reject or fail to reject the null hypothesis.
- Find the critical value for a test for correlation with α = 0.01 for a sample size of 18.The table below lists weights (carats) and prices (dollars) of randomly selected diamonds. Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. There is sufficient evidence o support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. For the prediction interval, use a 95% confidence level with a diamond that weighs 0.8 carats. Weight Price 0.3 $508 a. Find the explained variation. 0.4 $1153 0.5 $1332 Round to the nearest whole number as needed.) . Find the unexplained variation. Round to the nearest whole number as needed.) c. Find the indicated prediction interval.Unfortunately, arsenic occurs naturally in some ground water. A mean arsenic level of u = 8 parts per billion (ppb) is considered safe for agricultural use. A well in Los Banos is used to water cotton crops. This well is tested on a regular basis for arsenic. A random sample of 37 tests gave a sample mean of = 7.3ppb arsenic. It is known that o = 1.9 ppb for this type of data. Does this information indicate that the mean level of arsenic in this well is less than 8 ppb? Use the classical approach. Use a = 0.01 What is the Decision (step 5) for this problem? There is not sufficient evidence at the 0.01 level of significance to show the mean arsenic level in the Los Banos well is less than 7.3 ppb. O There is sufficient evidence at the 0.01 level of significance to show the mean arsenic level in the Los Banos well is less than 8.0 ppb. There is not sufficient evidence at the 0.01 level of significance to show the mean arsenic level in the Los Banos well is less than 8.0 ppb. O There is…In the year 2033, katy perry is a leading traveling nurse. Katy is interested in reducing the mean recovery time for patients after experiencing a serious injury (assume recovery times are normally distributed). Suppose the mean recovery time is presently 8.6 months. Katy takes a random sample of 46 patients that have experienced serious injury to participate in a new treatment program and finds the sample mean is 8.1 months and a sample standard deviation of 1.2 months. Using α = 0.05, answer the following questions. g) What is the interpretation (not your conclusions) of the p value in the context of the problem you found in part e? h) What are your conclusions in the context of the problem? Relate your conclusions to the test statistic and critical value, confidence interval and p value.A firm produces calculators. It manager claims that its calculators are good, on average, for at least 58 months. A consumer protection agency tested 16 such calculators to check this claim. He found the mean life of these calculators to be 55 months with a variance of 9 months. Can you conclude that the claim of this company is true? Use a = 0.05. Assume that the life of such a calculator has an approximately normal distribution. According to this information the null and alternative hypotheses are: H0: μ ≤ 58 H1: μ > 58 Select one: True FalseSuppose data is collected to predict y from x. If the size of Sample ✓ Calculate the p-value if the test statistic is t = -1.38IQ scores among the general population have a mean of 100 and a standard deviation of 14. A researcher claims that the standard deviation, o, of IQ scores for females is not equal to 14. A random sample of 24 IQ scores females had a mean of 98 and a standard deviation of 11. Assuming that IQ scores for females are approximately normally distributed, is there significant evidence (at the 0.05 level of significance) to conclude that the researcher's claim is correct? Perform a two-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified below. (If necessary, consult a list of formulas.) (a) State the null hypothesis H, and the alternative hypothesis H,. H, :0 H, :0 (b) Determine the type of test statistic to use. (Choose one) ▼ O=0 OSO (c) Find the value of the test statistic. (Round to three or more decimal places.) OA researcher predicts that smoking cigarettes decreases a person's sense of smell. On a test of olfactory sensitivity (smell), the mean for non-smokers is 18.4. A sample of 12 people who smoke a pack each day have a sample mean of 16.25 (sample variance = 4.75). Complete the following statement: We ["", ""] the null. Smokers have ["", "", ""] in olfactory sensitivity compared to non-smokers.The average score on a statistics test was 2.21 with a standard deviation of 0.88. The teacher suspects that the exam was very difficult; Based on the above, you want to adjust the scores so that the mean is 3.37 and the standard deviation is 0.25. Compute the values a and b such that the fit of type aX+b gives the desired mean and variance? Calculate: a= Answer b= Answer TIP: Use expected value and variance properties.SEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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