Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomlyselects a sample of 41 cookies and finds that the number of chocolate chips per cookie in the sample has amean of 10.7 and a standard deviation of 1.2. What is the 90% confidence interval for the number ofchocolate chips per cookie for Big Chip cookies? Enter your answers accurate to one decimal place (becausethe sample statistics are reported accurate to one decimal place).

The sample size n=41 The sample mean x¯=10.7 The sample standard deviation s=1.2 The confidence…

Answered: Karen wants to advertise how many…

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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The average McDonald's restaurant generates $3.6 million in sales each year with a standard deviation of 0.9. Trent wants to know if the average sales generated by McDonald's restaurants in Kentucky is different than the worldwide average. He surveys 27 restaurants in Kentucky and finds the following data (in millions of dollars): 4.1, 2.8, 4.4, 4.5, 5.3, 5, 3.7, 2.9, 3.8, 4.8, 3.6, 2.3, 3.7, 2.9, 2.9, 4, 1.1, 5.2, 2.9, 5, 4, 4, 5.9, 3.2, 2.2, 4.3, 3.8 Perform a hypothesis test using a 3% level of significance. Step 1: State the null and alternative hypotheses. Ho: [? v] ? v На: ? ? v (So we will be performing a Select an answer test.) Step 2: Assuming the null hypothesis is true, determine the features of the distribution of point estimates using the Central Limit Theorem. By the Central Limit Theorem, we know that the point estimates are Select an answer v with distribution mean and distribution standard deviation Step 3: Find the p-value of the point estimate. P( ? v ? v = P( ? ♥ ?…
A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 76.8. He knows the population standard deviation for the evening classes to be 7.2 points. A random sample of 200 students from morning classes results in a mean test score of 77.8. He knows the population standard deviation for the morning classes to be 1.9 points. Test his claim with a 90% level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ1=μ2 Ha: μ1__μ2
Tina catches a 14-pound bass. She does not know the population mean or standard deviation. So she takes a sample of five friends and they say the last bass they caught was 9, 12, 13, 10, and 10 pounds. Find the t and calculate a 95% (α = .05) confidence interval.
Lucy recently switched her primary doctor to one specializing in caring for elderly patients. On her new doctor's website, it says that the mean systolic blood pressure among elderly females is 120 millimeters of mercury (mmHg). Lucy believes the value is actually higher. She bases her belief on a recently reported study of 19 randomly selected, elderly females. The sample mean systolic blood pressure was 132 mmHg, and the sample standard deviation was 24 mmHg. Assume that the systolic blood pressures of elderly females are approximately normally distributed. Based on the study, at the 0.05 level of significance, can it be concluded that μ, the population mean systolic blood pressure among elderly females, is greater than 120 mmHg? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) esc (a) State the null hypothesis H and the alternative hypothesis H₁. Ho: H₁:0 (b)…
Ann recently switched her primary doctor to one specializing in caring for elderly patients. On her new doctor's website, it says that the mean systolic blood pressure among elderly females is 120 millimeters of mercury ( mmHg ). Ann believes the value is actually higher. She bases her belief on a recently reported study of 10 randomly selected, elderly females. The sample mean systolic blood pressure was 128 mmHg , and the sample standard deviation was 25 mmHg . Assume that the systolic blood pressures of elderly females are approximately normally distributed. Based on the study, at the 0.10 level of significance, can it be concluded that μ , the population mean systolic blood pressure among elderly females, is greater than 120 mmHg ? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) (a) State the null hypothesis H0 and the…
Heather recently switched her primary doctor to one specializing in caring for elderly patients. On her new doctor's website, it says that the mean systolic blood pressure among elderly females is 115 millimeters of mercury (mmHg). Heather believes the value is actually higher. She bases her belief on a recently reported study of 13 randomly selected, elderly females. The sample mean systolic blood pressure was 125 mmHg, and the sample standard deviation was 25 mmHg. C Assume that the systolic blood pressures of elderly females are approximately normally distributed. Based on the study, at the 0.05 level of significance, can it be concluded that u, the population mean systolic blood pressure among elderly females, is greater than 115 mmHg? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) (a) state the null hypothesis Ho and the alternative hypothesis H₁. 1 H₂ : 0…
John finds that there is a scholarship available to all persons scoring in the top 5 % on the ACT test. If the mean score on the ACT is 21 with a standard deviation of 4.7, what score does he need in order to qualify for the scholarship?
A sample of 134 motels is selected from a large urban area and the price for a night of lodging for a single room was determined for each motel. The mean rate is computed to be $87 and the standard deviation is $14. One motel charged $57 per night which is the 13th percentile. Another motel charged $118 per night which is the 75th percentile. Step 3 of 4 : One of the motels was 1.3 standard deviations below the mean, what is the price of a single room at the motel? Round your answer to two decimal places.
We would like to create a confidence interval for the mean. Which of the following would produce the smallest margin of error? Select one: a.A 95% confidence level and a sample size of 150 subjects b. A 99% confidence level and a sample size of 600 subjects. c. A 99% confidence level and a sample size of 150 subjects. d.A 95% confidence level and a sample size of 600 subjects.
A successful basketball player has a height of 6 feet 11 inches, or 211 cm. Based on statistics from a data set, his height converts to the z score of 5.17. How many standard deviations is his height above the mean?
A parenting magazine reports that the average amount of wireless data used by teenagers each month is 5 Gb. For her science fair project, Ella sets out to prove the magazine wrong. She claims that the mean among teenagers in her area is less than reported. Ella collects information from a simple random sample of 25 teenagers at her high school, and calculates a mean of 4.7 Gb per month with a standard deviation of 0.9 Gb per month. Assume that the population distribution is approximately normal. Test Ella's claim at the 0.01 level of significance. Step 3 of 3: Draw a conclusion and interpret the decision. E Tables E Keypad Answer Keyboard Shortcuts We reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance that the average amount of wireless data used by teenagers each month is less than 5 Gb. We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.01 level of significance that the average amount of…
A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 85.6. He knows the population standard deviation for the evening classes to be 4.6 points. A random sample of 150 students from morning classes results in a mean test score of 86.7. He knows the population standard deviation for the morning classes to be 8.3 points. Test his claim with a 99 % level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 3 of 3: Draw a conclusion and interpret the decision. Answer 围 Tables E Keypad Keyboard Shortcuts We fail to reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance to support the…

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