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f. The p-value for this sample= (Please show your answer to four decimal places.)

f. The p-value for this sample= (Please show your answer to four decimal places.)

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.4: Distributions Of Data
Problem 19PFA
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P12help part F please

A recent national report states the marital status distribution of the male population age 18 or older is as follows: Never Married (31.7%) , Married (54.6%), Widowed (2.4%), Divorced (11.3%). The table below shows the results of a random sample of 1872 adult men from California. Test the claim that the distribution from California is as expected at the = 0.05 significance level. a. Complete the table by filling in the expected frequencies. Round to the nearest whole number: Frequencies of Marital Status Outcome Frequency Expected Frequency Never Married 580 Married 1035 Widowed 27 O Divorced 230 212 b. What is the correct statistical test to use? Goodness-of-Fit ✔ O 593 1022 c. What are the null and alternative hypotheses? Ho: d. The degrees of freedom = 3 45 O Marital status and residency are independent. O The distribution of marital status in California is not the same as it is nationally. The distribution of marital status in California is the same as it is nationally. Marital status and residency are dependent. H₁: O The distribution of marital status in California is the same as it is nationally. ⒸThe distribution of marital status in California is not the same as it is nationally. Marital status and residency are independent. Marital status and residency are dependent. f. The p-value for this sample= 8 e. The test-statistic for this data = 9.179 (Please show your answer to three decimal places.) (Please show your answer to four decimal places.)
Transcribed Image Text:A recent national report states the marital status distribution of the male population age 18 or older is as follows: Never Married (31.7%) , Married (54.6%), Widowed (2.4%), Divorced (11.3%). The table below shows the results of a random sample of 1872 adult men from California. Test the claim that the distribution from California is as expected at the = 0.05 significance level. a. Complete the table by filling in the expected frequencies. Round to the nearest whole number: Frequencies of Marital Status Outcome Frequency Expected Frequency Never Married 580 Married 1035 Widowed 27 O Divorced 230 212 b. What is the correct statistical test to use? Goodness-of-Fit ✔ O 593 1022 c. What are the null and alternative hypotheses? Ho: d. The degrees of freedom = 3 45 O Marital status and residency are independent. O The distribution of marital status in California is not the same as it is nationally. The distribution of marital status in California is the same as it is nationally. Marital status and residency are dependent. H₁: O The distribution of marital status in California is the same as it is nationally. ⒸThe distribution of marital status in California is not the same as it is nationally. Marital status and residency are independent. Marital status and residency are dependent. f. The p-value for this sample= 8 e. The test-statistic for this data = 9.179 (Please show your answer to three decimal places.) (Please show your answer to four decimal places.)
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p17 need help with D F G please

A biologist looked at the relationship between number of seeds a plant produces and the percent of those seeds that sprout. The results of the survey are shown below. 51 Seeds Produced 68 67 66 Sprout Percent 60.8 57.2 55.6 68.6 a. Find the correlation coefficient: r = -0.83 b. The null and alternative hypotheses for correlation are: Ho: P✓ ✓ = 0 H₁: P م من 0+ The p-value is: 0.0056 43 74.8 (Round to four decimal places) c. Use a level of significance of ax = 0.05 to state the conclusion of the hypothesis test in the context of the study. 60 59 59 58 66 68.4 63.4 73.8 There is statistically significant evidence to conclude that there is a correlation between the number of seeds that a plant produces and the percent of the seeds that sprout. Thus, the regression line is useful. d. T² = 0.69 e. Interpret ²: Round to 2 decimal places. O There is statistically significant evidence to conclude that a plant that produces more seeds will have seeds with a lower sprout rate than a plant that produces fewer seeds. There is statistically insignificant evidence to conclude that a plant that produces more seeds will have seeds with a lower sprout rate than a plant that produces fewer seeds. There is statistically insignificant evidence to conclude that there is a correlation between the number of seeds that a plant produces and the percent of the seeds that sprout. Thus, the use of the regression line is not appropriate. x (Round to two decimal places) There is a large variation in the percent of seeds that sprout, but if you only look at plants that produce a fixed number of seeds, this variation on average is reduced by 68%. f. The equation of the linear regression line is: O Given any group of plants that all produce the same number of seeds, 68% of all of these plants will produce seeds with the same chance of sprouting. 68% of all plants produce seeds whose chance of sprouting is the average chance of sprouting. There is a 68% chance that the regression line will be a good predictor for the percent of seeds that sprout based on the number of seeds produced. (Please show your answers to two decimal places) g. Use the model to predict the percent of seeds that sprout if the plant produces 54 seeds. Percent sprouting = (Please round your answer to the nearest whole number.)
Transcribed Image Text:A biologist looked at the relationship between number of seeds a plant produces and the percent of those seeds that sprout. The results of the survey are shown below. 51 Seeds Produced 68 67 66 Sprout Percent 60.8 57.2 55.6 68.6 a. Find the correlation coefficient: r = -0.83 b. The null and alternative hypotheses for correlation are: Ho: P✓ ✓ = 0 H₁: P م من 0+ The p-value is: 0.0056 43 74.8 (Round to four decimal places) c. Use a level of significance of ax = 0.05 to state the conclusion of the hypothesis test in the context of the study. 60 59 59 58 66 68.4 63.4 73.8 There is statistically significant evidence to conclude that there is a correlation between the number of seeds that a plant produces and the percent of the seeds that sprout. Thus, the regression line is useful. d. T² = 0.69 e. Interpret ²: Round to 2 decimal places. O There is statistically significant evidence to conclude that a plant that produces more seeds will have seeds with a lower sprout rate than a plant that produces fewer seeds. There is statistically insignificant evidence to conclude that a plant that produces more seeds will have seeds with a lower sprout rate than a plant that produces fewer seeds. There is statistically insignificant evidence to conclude that there is a correlation between the number of seeds that a plant produces and the percent of the seeds that sprout. Thus, the use of the regression line is not appropriate. x (Round to two decimal places) There is a large variation in the percent of seeds that sprout, but if you only look at plants that produce a fixed number of seeds, this variation on average is reduced by 68%. f. The equation of the linear regression line is: O Given any group of plants that all produce the same number of seeds, 68% of all of these plants will produce seeds with the same chance of sprouting. 68% of all plants produce seeds whose chance of sprouting is the average chance of sprouting. There is a 68% chance that the regression line will be a good predictor for the percent of seeds that sprout based on the number of seeds produced. (Please show your answers to two decimal places) g. Use the model to predict the percent of seeds that sprout if the plant produces 54 seeds. Percent sprouting = (Please round your answer to the nearest whole number.)
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