### Using the Standard Normal Distribution#### Steps to Conduct a Test:1. **Select one-tailed or two-tailed:** - Options: - One-tailed - Two-tailed2. **Enter the test statistic:** - Note: Round to 3 decimal places.3. **Shade the area represented by the *p*-value:** - This step involves visualizing the area under the curve corresponding to the *p*-value.4. **Enter the *p*-value:** - Note: Round to 3 decimal places.#### Graph Details:The graph shown is a standard normal distribution curve. It is symmetric and bell-shaped with the mean at the center (0) and standard deviations marked along the x-axis. The y-axis represents the probability density.#### Decision Making Based on Hypothesis Testing:**(c) Based on your answer to part (b), determine the conclusion at the 0.10 level of significance:**- **Options:** 1. Since the *p*-value is less than (or equal to) the level of significance, the null hypothesis is rejected. Thus, there is enough evidence to support the claim that the mean customer rating is not equal to 79. 2. Since the *p*-value is less than (or equal to) the level of significance, the null hypothesis is not rejected. Thus, there is not enough evidence to support the claim that the mean customer rating is not equal to 79. 3. Since the *p*-value is greater than the level of significance, the null hypothesis is rejected. Thus, there is enough evidence to support the claim that the mean customer rating is not equal to 79. 4. Since the *p*-value is greater than the level of significance, the null hypothesis is not rejected. Thus, there is not enough evidence to support the claim that the mean customer rating is not equal to 79.**Buttons:**- Explanation- Check© 2021 McGraw Hill LLC. All Rights Reserved. Over the years, the mean customer satisfaction rating at a local restaurant has been 79. The restaurant was recently remodeled, and now the management claims the mean customer rating, μ, is not equal to 79. In a sample of 51 customers chosen at random, the mean customer rating is 79.3. Assume that the population standard deviation of customer ratings is 2.4.Is there enough evidence to support the claim that the mean customer rating is different from 79? Perform a hypothesis test, using the 0.10 level of significance.(a) State the null hypothesis $ H_0 $ and the alternative hypothesis $ H_1 $:- $ H_0 : \mu = 79 $- $ H_1 : \mu \neq 79 $(b) Perform a Z-test and find the p-value. Here is some information to help you with your Z-test:- The value of the test statistic is given by $ \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} $- The p-value is two times the area under the curve to the right of the value of the test statistic.**Graph Description:**The graph displays a standard normal distribution curve. A vertical line represents the Z-test statistic's value, which separates the curve into two tails. The shaded area to the right indicates the p-value for a two-tailed test.**Steps:**1. **Select one-tailed or two-tailed:** - One-tailed - Two-tailed (selected for this test)2. **Enter the test statistic:** (Round to 3 decimal places)**Note:**The test involves assessing whether the customer's mean satisfaction rating has significantly changed from the historical mean of 79, considering an alpha level of 0.10 for the hypothesis test.

Answered: Over the years, the mean customer…

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 pharmaceutical company needs to know if its new cholesterol drug, Praxor, is effective at lowering cholesterol levels. It believes that people who take Praxor will average a greater decrease in cholesterol level than people taking a placebo. After the experiment is complete, the researchers find that the 46 participants in the treatment group lowered their cholesterol levels by a mean of 19.5 points with a standard deviation of 4.9 points. The 37 participants in the control group lowered their cholesterol levels by a mean of 18.4 points with a standard deviation of 4.1 points. Assume that the population variances are not equal and test the company's claim at the 0.02 level. Let the treatment group be Population 1 and let the control group be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.
The average credit card debt for a recent year was $8,776. Five years earlier the average credit card debt was $8,189. Assume sample sizes of 32 were used and the population standard deviations of both samples were $690. Is there evidence to conclude that the average credit card debt has increased?
Salma 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). Salma believes the value is actually higher. She bases her belief on a recently reported study of 21 randomly selected, elderly females. The sample mean systolic blood pressure was 118 mmH, and the sample standard deviation was 21 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 y, 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.)
It is commonly believed that the mean body temperature of a healthy adult is 98.6°F. You are not entirely convinced. You believe that it is not 98.6° F. You collected data using 52 healthy people and found that they had a mean body temperature of 98.25° F with a standard deviation of 1.2° F. Use a 0.05 significance level to test the claim that the mean body temperature of a healthy adult is not 98.6°F. a) kdentify the null and alternative hypotheses? Họ:p=v 98.6 H.: p
According to the Insurance Association in US, the average annual expenditure for automobile insurance is $605. Suppose that in a simple random sample of 80 residents of New York, for most recent year their average expenditure was $638.62 with the standard deviation of $106.05. Based on these data, examine whether the average annual insurance for motorists in New York might be different from that in national level. Using the 0.05 level of significance, what conclusion do you reach? Also, construct and interpret 95% confidence interval for the population mean. Is the hypothesized mean within the interval? Is this consistent with the findings of the hypothesis test conducted?
A random sample of 64 shoppers showed that they spend an average of $27.31 per visit at the Sunday Mornings Bookstore. The standard deviation of the population is $2.14. Find the 90% confidence interval of the true mean. Please include all work showing how you found your answer.
Christin ently 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). Christine believes the value is actually higher. She bases her belief on a recently reported study of 14 randomly selected, elderly females. The sample mean systolic blood pressure was 129 mmHg, and the sample standard deviation was 22 Español 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 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 H and the alternative hypothesis H . H, :0…
In a sample of 351 gamers between the ages of 16-25, the mean number of hours spent gaming per week was 42.5 hours. In a sample of 269 gamers between the ages of 26-40, the mean number of hours spent gaming per week was 43.8 hours. If the population standard deviation for the mean number of hours spent gaming per week is 9.5 for the age group 16-25 and 10.7 for the age group 26-40, test the claim that gamers between the ages of 16-25 spend less time gaming per week than gamers between the ages of 26-40. Assume α=0.03. Let population 1 be gamers in the age group 16-25 and population 2 be gamers in the age group 26-40. Round the test statistic to two decimal places, and round the p-value to four decimal places. The test statistic is . The p-value is . Should the null hypothesis be rejected?
In a certain city, the average 20- to 29-year old man is 72.5 inches tall, with a standard deviation of 3.2 inches, while the average 20- to 29-year old woman is 64.3 inches tall, with a standard deviation of 3.9 inches. Who is relatively taller, a 75-inch man or a 70-inch woman?
A college in the northeastern United States used SAT scores to help decide which applicants to admit. To determine whether the SAT was useful in predicting success, the college examined the relationship between the SAT scores and first-year GPAs of admitted students. The average math SAT score of a sample of 200 randomly selected students was 649.5 with a standard deviation of 66.2. The average GPA of the same students was 2.63 with a standard deviation of 0.58. The correlation between GPA and math SAT score was 0.19. (a) What does the correlation of 0.19 tell us about SAT scores and GPA scores? (You need to provide two answers.)(b) Find the equation of the least squares regression line.(c) Using the equation, predict the GPA score of a person with a math SAT score of 650.(d) Using the equation, predict the GPA score for a person with a math score of 800.
A pharmaceutical company needs to know if its new cholesterol drug, Praxor, is effective at lowering cholesterol levels. It believes that people who take Praxor will average a greater decrease in cholesterol level than people taking a placebo. After the experiment is complete, the researchers find that the 34 participants in the treatment group lowered their cholesterol levels by a mean of 22.2 points with a standard deviation of 3.4 points. The 42 participants in the control group lowered their cholesterol levels by a mean of 21.2 points with a standard deviation of 1.8 points. Assume that the population variances are not equal and test the company's claim at the 0.10 level. Let the treatment group be Population 1 and let the control group be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho:₁ = ₂ Ha: M •M₂
A study is done to determine which of two soft drinks has more sugar. There are 25 cans of Beverage A in the researchers' sample and 15 cans of Beverage B. The mean amount of sugar in Beverage A is 34.7 grams per can with a sample standard deviation of 0.7 grams. The mean amount of sugar in Beverage B is 33.9 grams with a standard deviation of 0.6 grams. The researchers believe that Beverage A has less sugar than Beverage B, on average. Both populations are normally distributed. Conduct a hypothesis test using a 2% level of significance. Step 1: State the null and alternative hypotheses. Ho:µA – µB 2V Ho:HA - HB V test.) (So we will be performing a left-tailed forum

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