A company claims that each batch of its deluxe mixed nuts contains 52% cashews, 27% almonds, 13% macadamia nuts, and 8%Brazil nuts. To test this claim, a quality-control inspector takes a random sample of 150 nuts from the latest batch. The tabledisplays the sample data.Type of nut Cashew Almond Macadamia BrazilCount8329201819.5,The inspector has also calculated the expected counts for each type of nut: Cashew = 78, Almond = 40.5, Macadamia =Brazil = 12.The Random condition is met. Can the inspector use a chi-square distribution to calculate the P-value?Because at least one expected count is greater than 30, the inspector can use a chi-square distribution.Because all expected counts are not greater than 30, the inspector cannot use a chi-square distribution.Because all observed counts are greater than 5, the inspector can use a chi-square distribution.Because the distribution of counts is approximately normal, the inspector can use a chi-square distribution.Because all expected counts are greater than 5, the inspector can use a chi-square distribution.

The given expected counts are: Type of nut Cashew Almond Macadamia Brazil Expected count 78…

Answered: A company claims that each batch of its…

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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Similar questions

About 4% of the population has a particular genetic mutation. 300 people are randomly selected.Find the mean for the number of people with the genetic mutation in such groups of 300. (Remember that means should be rounded to one more decimal place than the raw data.)
A researcher wants to investigate whether the education level is related to his or her place of residence. Hundred's people were selected randomly. The results are given as follow. Area * Education Level Crosstabulation Education Level Bachelor's SPM Diploma Degree Total Area Urban Count 22 10 18 50 Expected Count 25.0 8.0 17.0 50.0 Rural Count 28 16 50 Expected Count 25.0 8.0 A 50.0 Total Count 50 16 34 100 Expected Count 50.0 16.0 34.0 100.0 Chi-Square Tests Value df Pearson Chi-Square в 2 Likelihood Ratio 1.850 2 Linear-by-Linear Association .778 1 N of Valid Cases a. O cells (0.0%) have expected count less than 5. The 100 minimum expected count is 8.00. a) State a reason of choosing the Chi-Square Test of Independence for this study. b) Calculate the values of A and B. c) At the 5% significance level, test whether education level is associated to his or her place of residence.
A market research firm has carried out an observational study on the chief executives from a random selection of the law firms in the country. The numerical measures obtained from the analysis of the data are referred to as ______? a) records b) statistics c) parameters d) variables
International experience around the world is important in our culture of globalization. According to NAFSA, 54.5% of all American students traveling abroad went to Europe in the 2014-2015 academic year. To determine if there is shift in the proportion of students traveling to Europe, suppose a random sample of 300 students who had studied abroad was taken for the 2015-2016 school year. One-hundred sixty of the students traveled to Europe. a. Conduct a test to determine if the proportion of students traveling to Europe has changed. Be sure to provide a complete write-up. b. Based on the hypothesis test, would you expect 0.545 to be in confidence interval for the proportion? Explain without constructing the interval.
The city of Hampton is surveying households to see if the number of cats per household has changed over the past decade. The last known average of cats per household was 1.08. What statistical procedure should you use to analyze these data? O one sample z test Oone sample t test O correlated groups t test O independent groups t test
From a Population consisting of following values of 2,2,4,4 and 8, how many samples of size two can be drawn ? Select one: a.12 b.10 c.13 d.9 e.11
A certain virus affects 0.7% of the population. A test used to detect the virus in a person is positive 87% of the time if the person has the virus (true positive) and 14% of the time if the person does not have the virus (false positive). Fill out the remainder of the following table and use it to answer the two questions below based on a total sample of 100,000 people. Virus No Virus TotalPositive Test Negative Test Total 100,000a) Find the probability that a person has the virus given that they have tested positive. Round your answer to the nearest hundredth of a percent and do not include a percent sign. % b) Find the probability that a person does not have the virus given that they test negative. Round your answer to the nearest hundredth of a percent and do not include a percent sign. %
A governmental agency computed the proportion of violent crimes in the United States in a particular year falling into each of four categories. A simple random sample of 500 violent crimes committed in California during that year were categorized in the same way. The following table presents the results. Category U.S.Proportion CaliforniaFrequency Murder 0.018 4 Forcible Rape 0.07 35 Robbery 0.390 218 Aggravated Assault 0.522 243 Can you conclude that the proportions of crimes in the various categories in California differ from those in the United States as a whole? Use the 0.05 level of significance and the P-value method with the TI-84 Plus calculator. (b) Find the P-value. Round the answer to at least four decimal places.
Consider these sample data: x, = 12, x2 = 7, X3 = 6, x4 = 22. a. Find n. b. Compute x. c. Determine x.
According to the NIMH website the average number of adult Americans diagnosed with a “serious mental illness” across all 50 states in 2019 was μ = 5.2 out of 100 residents (or 5.2%). Please test to see if the sample of states listed below (n=8) has a significantly higher level of mental illness than the population mean of 5.2. Sample data: Arkansas= 5.45%, Montana=5.34%, Vermont=5.32%, Oregon=5.19%, Utah=5.2%, Ohio=5.13, Kentucky=5.25% West Virginia=5.18%. Hint - drop the % signs and use 5.45, 5.32, etc as the X scores. If you want you can use JASP for help with the calculations. Use 3 decimal places for the M, s, SEM. The alpha level for the study was set to α=.05. a. State the hypotheses of the study. b. Find the critical t value for this studyc. Compute the one-sample t test. d. Compute the Cohen's d for the data (even if not significant) e. State the results - is the sample significantly higher? Do you reject the null hypothesis? Report the t test results in APA format…
According to the M&M candy company, the expected proportion can be found in the table. In addition, the table contains the number of M&M's of each color that were found in a case of candy. At the 10% level, do the observed frequencies support the claim of M&M? M&M Observed and Proportions Brown Green Blue Orange Red Yellow Total 474 540 333 381 2734 Observed Frequencies 642 364 0.24 0.13 0.16 0.20 Expected Proportion 0.13 0.14 State the hypotheses. Ho: The observed frequencies are consistent with the distribution of colors for M&M candies. consistent with the distribution of colors for M&M candies. Ha: The observed frequencies are not Calculate the x² test statistic. Round expected values to two decimal places. Round x² to three decimal places. x² = Calculate the p-value. Round to four decimal places. p-value = State your decision. Since p-value?a, Select an answer Ho. Interpret the results. At the 10% signficance level there Select an answer enough evidence to conclude that the…
A study was conducted to determine how people get jobs. The table list date from 400 randomly selected subjects. The data are bases on results from the National centter for Career Strategies. Job source of survey respondent FREQUENCIES HELP-WANTED ads 56 EXECUTIVE SEARCH FIRMS 44 NETWORKING 280 MASS MAILING 20 TOTAL 1. What percentage of the respondents got job by either help-wanted ads or mass mailing? 2. What percentage of the respondent got jobs from networking?

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