John works as a weld inspector in a shipyard. Because he keeps track of good and poor welds, heknows that 5% of all welds will be poor during the afternoon shift. What are the chances that theproportion of poor welds will be more than 0.083 based on a sample of 300 welds?O a 0.0044Ob 0.9957Ⓒc 0.0045Od 0.9881

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Answered: John works as a weld inspector in a…

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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Bighorn sheep are found in the mountainous western United States. In the following data, x = the age of bighorn sheep in years and y = the percentage of that age group that die during the course of a year (mortality rate). In the table below, when x = 5, y = 20 meaning that 20% of the sheep aged between 5 and 6 years old die. A random sample of sheep provided the following information: x 1 2 3 4 5 y 14 18.9 14.4 19.6 20 ∑ x = 15 ; ∑ y = 86.9 ; ∑ x 2 = 55 ; ∑ y 2 = 1544.73 ; ∑ x y = 273.4 find m for this data
A random sample of n1 = 10 regions in New England gave the following violent crime rates (per million population). x1: New England Crime Rate 3.3 3.9 4.2 4.1 3.3 4.1 1.8 4.8 2.9 3.1 Another random sample of n2 = 12 regions in the Rocky Mountain states gave the following violent crime rates (per million population). x2: Rocky Mountain Crime Rate 3.7 4.1 4.7 5.5 3.3 4.8 3.5 2.4 3.1 3.5 5.2 2.8 Assume that the crime rate distribution is approximately normal in both regions. Do the data indicate that the violent crime rate in the Rocky Mountain region is higher than in New England? Use ? = 0.01. Solve the problem using both the traditional method and the P-value method. (Test the difference ?1 − ?2. Round the test statistic and critical value to three decimal places.) test statistic critical value Find (or estimate) the P-value. P-value > 0.250 0.125 < P-value < 0.250 0.050 < P-value < 0.125 0.025 < P-value < 0.050 0.005 <…
A random sample of n1 = 10 regions in New England gave the following violent crime rates (per million population). x1: New England Crime Rate 3.3 3.9 4.2 4.1 3.3 4.1 1.8 4.8 2.9 3.1 Another random sample of n2 = 12 regions in the Rocky Mountain states gave the following violent crime rates (per million population). x2: Rocky Mountain Crime Rate 3.7 4.1 4.7 5.5 3.3 4.8 3.5 2.4 3.1 3.5 5.2 2.8 Assume that the crime rate distribution is approximately normal in both regions. Do the data indicate that the violent crime rate in the Rocky Mountain region is higher than in New England? Use ? = 0.01. Solve the problem using both the traditional method and the P-value method. (Test the difference ?1 − ?2. Round the test statistic and critical value to three decimal places.) test statistic critical value Find (or estimate) the P-value. P-value > 0.2500.125 < P-value < 0.250 0.050 < P-value < 0.1250.025 < P-value < 0.0500.005 <…
The first four deviations from the mean in a sample of n = 5 reaction times were 0.1, 0.5, 1.1, and 1.5. What is the fifth deviation from the mean? Give a sample for which these are the five deviations from the mean. О -3.1, -2.7, -2.1, -1.7, -3.2 O 1.1, -0.5, 2.1, 2.5, -4.2 О 3.3, 3.7, 4.3, 4.7, 0 О -0.9, -0.5, 0.1, 0.5, 0
A random sample of n1 = 10 regions in New England gave the following violent crime rates (per million population). x1: New England Crime Rate 3.3 3.7 4.2 3.9 3.3 4.1 1.8 4.8 2.9 3.1 Another random sample of n2 = 12 regions in the Rocky Mountain states gave the following violent crime rates (per million population). x2: Rocky Mountain Crime Rate 3.5 4.1 4.7 5.5 3.3 4.8 3.5 2.4 3.1 3.5 5.2 2.8 Assume that the crime rate distribution is approximately normal in both regions. Do the data indicate that the violent crime rate in the Rocky Mountain region is higher than in New England? Use ? = 0.01. Solve the problem using both the traditional method and the P-value method. (Test the difference ?1 − ?2. Round the test statistic and critical value to three decimal places.) test statistic critical value Find (or estimate) the P-value. A. P-value > 0.250 B. 0.125 < P-value < 0.250 C. 0.050 < P-value < 0.125 D. 0.025 < P-value <…
A random sample of n1 = 10 regions in New England gave the following violent crime rates (per million population). x1: New England Crime Rate 3.3 3.9 4.2 4.1 3.3 4.1 1.8 4.8 2.9 3.1 Another random sample of n2 = 12 regions in the Rocky Mountain states gave the following violent crime rates (per million population). x2: Rocky Mountain Crime Rate 3.5 4.1 4.5 5.5 3.3 4.8 3.5 2.4 3.1 3.5 5.2 2.8 Assume that the crime rate distribution is approximately normal in both regions. Do the data indicate that the violent crime rate in the Rocky Mountain region is higher than in New England? Use ? = 0.01. Solve the problem using both the traditional method and the P-value method. (Test the difference ?1 − ?2. Round the test statistic and critical value to three decimal places.)
Q5. Borer et al. [1980], study 45 individuals after an acute myocardial infarction (heart attack). They measure the ejection fraction (EF); the EF is the percent of the blood pumped from the left ventricle (the pumping chamber of the heart) during a heart beat. A low EF indicates damaged or dead heart muscle (myocardium). During follow-up four patients died. Dividing EF into low ( 35% 4. 9. 0. 32 Is there reason to suspect, at a 0.05 significance level, that death is more likely in the low EF group? Use a one-sided p-value for your answer, since biological plausibility (and prior literature) indicates low EF is a risk factor for mortality.
We take a simple random sample (SRS) from a population to study the effect of marital status (married vs. not married on systolic blood pressure. Which t procedure should we use? A. a one-sample t procedureB. a paired t procedureC. ANOVAD. Noe of the above
Many college and university students obtain summer jobs. A professor wanted to determine whether students in different degree programs earn different amounts. She randomly selected five students in each of the 3 university programs and asked them to report what they earned the previous summer (in thousands of dollars). The following partially filled-in ANOVA table was obtained. Fill-in the rest of the table. (Use exact values for df and round all other numbers to 2 decimal places, if needed.) Source df MS F-Stat Treatments 12.4 Error Total 23.5
Have you ever been frustrated because you could not get a container of some sort to release the last bit of its contents? An article reported on an investigation of this issue for various consumer products. Suppose five 6.0 oz tubes of toothpaste of a particular brand are randomly selected and squeezed until no more toothpaste will come out. Then each tube is cut open and the amount remaining is weighed, resulting in the following data: 0.53, 0.64, 0.41, 0.5, 0.35. Does it appear that the true average amount left is less than 10% of the advertised net contents?Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)t= P-value =
(iii) Find (or estimate) the P-value. OP-value > 0.250 O 0.125 < P-value < 0.250 O 0.050 < P-value < 0.125 O 0.025 < P-value < 0.050 O 0.005< P-value < 0.025 OP-value < 0.005 Sketch the sampling distribution and show the area corresponding to the P-value. O -4 -2 0 2 2 M 2 DO Q -2 -2 (iv) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level a? O At the a= 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant. O At the a= 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. O At the a= 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant. O At the a= 0.01 level, we reject the null hypothesis and conclude the data are statistically significant. (v) Interpret your conclusion in the context of the application. O Fail to reject the null hypothesis, there…
a population proportion of 0.61. Suppose a random sample of 662 items is sampled randomly from this population

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