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For Problems 9-19, please provide the following information.
(a) What is the level of significance? State the null and alternate hypotheses.
(b) Find the value of the chi-square statistic for the sample. Are all the expected frequencies greater than 5? What sampling distribution will you use? What are the degrees of freedom?
(c) Find or estimate the P-value of the sample test statistic.
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?
(e) Interpret your conclusion in the context of the application.
Use the
Psychology: Myers-Briggs The following table shows the Myers-Briggs personality preferences for a random sample of 519 people in the listed professions (Atlas of Type Tables by Macdaid, McCaulley, and Kainz). T refers to thinking, and F refers to feeling.
Personality Preference Type | |||
Occupation | T | F | Row Total |
Clergy (all denominations) | 57 | 91 | 148 |
MD | 77 | 82 | 159 |
Lawyer | 118 | 94 | 212 |
Column Total | 252 | 267 | 519 |
Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.01 level of significance.
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Chapter 11 Solutions
EBK UNDERSTANDING BASIC STATISTICS
- I need help with this problem and an explanation of the solution for the image described below. (Statistics: Engineering Probabilities)arrow_forwardI need help with this problem and an explanation of the solution for the image described below. (Statistics: Engineering Probabilities)arrow_forwardDATA TABLE VALUES Meal Price ($) 22.78 31.90 33.89 22.77 18.04 23.29 35.28 42.38 36.88 38.55 41.68 25.73 34.19 31.75 25.24 26.32 19.57 36.57 32.97 36.83 30.17 37.29 25.37 24.71 28.79 32.83 43.00 35.23 34.76 33.06 27.73 31.89 38.47 39.42 40.72 43.92 36.51 45.25 33.51 29.17 30.54 26.74 37.93arrow_forward
- I need help with this problem and an explanation of the solution for the image described below. (Statistics: Engineering Probabilities)arrow_forwardSales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 65 weekly reports showed a sample mean of 19.5 customer contacts per week. The sample standard deviation was 5.2. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel. 90% Confidence interval, to 2 decimals: ( , ) 95% Confidence interval, to 2 decimals:arrow_forwardA simple random sample of 40 items resulted in a sample mean of 25. The population standard deviation is 5. a. What is the standard error of the mean (to 2 decimals)? b. At 95% confidence, what is the margin of error (to 2 decimals)?arrow_forward
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- College Algebra (MindTap Course List)AlgebraISBN:9781305652231Author:R. David Gustafson, Jeff HughesPublisher:Cengage LearningGlencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw Hill
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