**Educational Text and Explanation of Graphs**You have taken a random sample of size $ n = 19 $ from a normal population that has a population mean of $ \mu = 50 $ and a population standard deviation of $ \sigma = 15 $. Your sample, which is Sample 1 in the table below, has a mean of $ \bar{x} = 45.3 $.### Instructions:**(a)** Based on Sample 1, graph the 75% and 90% confidence intervals for the population mean. Use 1.150 for the critical value for the 75% confidence interval, and use 1.645 for the critical value for the 90% confidence interval.- Enter the lower and upper limits on the graphs to show each confidence interval. Write your answers with one decimal place.- For the points ($\color{red} \bullet$ and $\color{blue} \bullet$), enter the population mean, $\mu = 50$.**Graphical Explanation:**1. **75% Confidence Interval:** - A line is shown with markers from approximately 37.0 to 66.0, centering around a mean value for the sample ($45.3$), with a marker at 51.5 indicating the interval center.2. **90% Confidence Interval:** - Similarly, a line shows values from approximately 37.0 to 66.0, with a marker at 51.5 for the interval center.**(b)** Press the "Generate Samples" button to simulate taking 19 more samples of size $ n = 19 $ from the population. Notice that the confidence intervals for these samples are drawn automatically. Then complete parts (c) and (d) below the table.**Table and Additional Graphs:**The table shows sample means ($S1$ to $S20$) and placeholders for confidence interval limits. The columns indicate:- Each sample's mean- The lower and upper limits for both 75% and 90% confidence intervals**Graphical Explanation:**- The left side shows a vertical array of the 75% confidence intervals for each sample.- The right side shows a vertical array of the 90% confidence intervals.The graphs demonstrate how each sample’s confidence interval is calculated and illustrates the distribution of sample means around the population mean ($ \ (c) Notice that for \( \frac{16}{20} = 80\% $ of the samples, the 90% confidence interval contains the population mean. Choose the correct statement.- ○ When constructing 90% confidence intervals for 20 samples of the same size from the population, at most 90% of the samples will contain the population mean.- ○ When constructing 90% confidence intervals for 20 samples of the same size from the population, exactly 90% of the samples must contain the population mean. There must have been an error with the way our samples were chosen.- ○ When constructing 90% confidence intervals for 20 samples of the same size from the population, the percentage of the samples that contain the population mean should be close to 90%, but it may not be exactly 90%.(d) Choose ALL that are true.- □ The 75% confidence interval for Sample 15 indicates that 75% of the Sample 15 data values are between 44.4 and 52.4.- □ From the 90% confidence interval for Sample 15, we cannot say that there is a 90% probability that the population mean is between 42.7 and 54.1.- □ If there were a Sample 21 of size $ n = 38 $ taken from the same population as Sample 15, then the 90% confidence interval for Sample 21 would be narrower than the 90% confidence interval for Sample 15.- □ The 75% confidence interval for Sample 15 is narrower than the 90% confidence interval for Sample 15. This is coincidence; when constructing a confidence interval for a sample, there is no relationship between the level of confidence and the width of the interval.- □ None of the choices above are true.

Here we have give that,sample size = n = 19Population Mean = µ = 50Population Standard Deviation = σ…

Answered: You have taken a random sample of size…

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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For data that is not normally distributed we can't use z-scores. However, there is an equation that works on any distribution. It's called Chebyshev's formula. The formula is p = 1 - 7.2 where p is the minimum percentage of scores that fall within k standard deviations on both sides of the mean. Use this formula to answer the following questions. a) If you have scores and you don't know if they are normally distributed, find the minimum percentage of scores that fall within 2.1 standard deviations on both sides of the mean? b) If you have scores that are normally distributed, find the percentage of scores that fall within 2.1 standard deviations on both sides of the mean? c) If you have scores and you don't know if they are normally distributed, how many standard deviations on both sides of the mean do we need to go to have 96 percent of the scores? Note: To answer part c you will need to solve the equation for k.
Suppose that replacement times for washing machines are normally distributed with a mean of μ= 8.7 years and a standard deviation of σ= 1.6 years. Find and graph P25 the replacement time that separates the bottom 25% from the top 75%.
A researcher obtained the following results for a study: n = 44, SSy = 340603, SSx = 33712, B̂ = 2.0025 Find the correlation for these results?
While performing searches on the Internet to find accommodation, Jasmine realized that the cost of a housing unit seems to be related to the number of inhabitants of the municipality where the unit is located. She noted the following data. She represented the data in a scatter plot, where she drew a line of best fit passing through the mean point of the distribution. If Jasmine considers that a linear correlation exists between these two variables, what should be the mean monthly rent of a 4½-room unit in a municipality with a population of approximately 250 000? Justify your answer algebraically.
#2) A large warehouse installed a security system a few years ago. Under an older system the managers estimated that they were losing an average of $678 worth of goods to thieves each week. A random sample of 36 weekly records under the current system showed that they were still losing an average of $650 worth of goods each week. The sample standard deviation was s = $93. Does this indicate that the average loss is different (either more or less) than the previous $678 per week? Use a 5% level of significance.
A set of n=15 pairs of scores (x and y values) has Ssx =200, Ssy=60, and Sp=30.What is the Pearson correlation for these data?
A standardized test is given to a sixth-grade class. Historically the mean score has been 151 with a standard deviation of 21. The superintendent believes that the standard deviation of performance may have recently decreased. She randomly sampled 23 students and found a mean of 161 with a standard deviation of 18.2898. Is there evidence that the standard deviation has decreased at the α=0.05 level? Step 1 of 5 : State of the hypotheses in terms of the standard deviation. Round the standard deviation to four decimal places when necessary Step 2 of 5 : Determine the critical value(s) of the test statistic. If the test is two-tailed, separate the values with a comma. Round your answer to three decimal places. Step 3 of 5 : Determine the value of the test statistic. Round your answer to three decimal places. Step 4 of 5 : Make the decision. Step 5 of 5 : What is the conclusion?
#10. A psychologist is studying the self image of smokers, which she measures by the self image (SI) score from a personality inventory. She would like to estimate the mean SI score, U ( mean population), for the population of all smokers. She plans to take a random sample of SI scores for smokers an estimate U (population mean) via the sample. Assuming that the standard deviation of SI scores for the population of all smokers is 82, what is the minimum sample size needed for the psychologist to be 99% confident that her estimate is within 12 of U (population mean)? Carry your intermediate computations to at least three decimal places write your answer as a whole number, and make sure it is the minimum whole number that satisfies the requirements .

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