A First Course in Probability (10th Edition)
10th Edition
ISBN: 9780134753119
Author: Sheldon Ross
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Question
Nationwide, suppose that drinking water samples have probability 0.04 of having insufficient levels of chlorine. Suppose further that a recent study in alberta showed that 12 of 500 drinking water samples had insufficient levels of chlorine. Does this evidence imply that alberta is significantly under the national average? Test at a 5% significance level and report a p-value.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by stepSolved in 2 steps with 2 images
Knowledge Booster
Similar questions
- In a previous poll, 46% of adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Suppose that, in a more recent poll, 515 of 1172 adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Is there sufficient evidence that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has decreased? Use the a= 0.05 significance level. Because npo (1- Po) = V 10 and the sample size is V 5% of the population size, and the sample v the requirements for testing the hypothesis v satisfied. (Round to one decimal place as needed.) What are the null and alternative hypotheses? Ho: versus H, (Type integers or decimals. Do not round.) Find the test statistic, Zo- Zo (Round to two decimal places as needed.) Find the P-value. P-value = (Round to three decimal places as needed.) Is there sufficient evidence that the proportion of families with children…arrow_forwardSuppose IQ scores were obtained for 20 randomly selected sets of siblings. The 20 pairs of measurements yield x = 98.67, y = 99, r= 0.930, P-value = 0.000, and %3D y = - 12.38 +1.13x, where x represents the IQ score of the younger child. Find the best predicted value of y given that the younger child has an IQ of 97? Use a significance level of 0.05. Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted value of y is (Round to two decimal places as needed.)arrow_forwardCadmium, a heavy metal, is toxic to animals. Mushrooms, however, are able to absorb and accumulate cadmium at high concentrations. A government set a safety limit for cadmium in dry vegetables at 0.5 parts per million (ppm). The cadmium levels in a random sample of one species of edible mushroom are in the accompanying data set. At the 10% significance level, do the data provide sufficient evidence to conclude that the mean cadmium level in this species of mushroom is greater than the government's recommended limit of 0.5ppm? Assume that the population standard deviation of cadmium levels in this species of mushroom is 0.45 ppm. Preliminary data analyses indicate that applying the z-test is reasonable. (Note: The sum of the data is 6.57 ppm.) 0.32 0.44 0.17 0.23 0.79 1.21 0.84 0.24 0.46 0.41 0.92 0.54 State the hypotheses for the one-mean z-test.H0: μ (≠, >, <, =) ________ppmHa: μ (≠, >, <, =) _________ppmCompute the value of the test…arrow_forward
- Suppose IQ scores were obtained for 20 randomly selected sets of siblings. The 20 pairs of measurements yield x = 101.13, y = 103.5, r=0.867, P-value = 0.000, and y = 6.25 +0.96x, where x represents the IQ score of the younger child. Find the best predicted value of y given that the younger child has an IQ of 101? Use a significance level of 0.05. Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted value of ŷ is. (Round to two decimal places as needed.)arrow_forwardSuppose a study reported that the average persin watched 3.37 hours of television per day. a random sample of 15 people gave the number of hours of television watched per day shown below. at the 1% significance level, do the data provide sufficent evidence to conclude that the amount of television watched per day last year by the average person is greater than the value reported in the study? what is the test statistic value for this problem? a.0.28 b.-38 c.none of the above d. 0.58 e.-4.295 f..04 g. 1.94 . the proper conclusion for this problem is? a. reject the null hypothesis we have sufficent evidence to prove the average number of hours people watch tv is more than 3.37 hours. b. Do not reject the null hypothesis we have sufficient evidence to prove the average number of hours of TV watched is greater than 3.37 hours c. Reject the null hypothesis and claim the average number of hours people watch TV is less than 3.37 because the P value is near zero d. Do not reject the null…arrow_forwardThe national percentage of automobile accident fatalities that are alcohol related is 39%. In a random sample of 96 automobile accident fatalities in the state of New York, 44 were alcohol related. Is this sufficient evidence to say that new York has a higher percentage of alcohol related automobile fatalities? a. Perform the appropriate test of hypothesis and paste the session window output below. b. What is the value of the test statistic? c. At a = .10, what is your decision?arrow_forward
- A company claims that at least 94% of equipment supplied to a manufacturer is within specifications. A test of a sample of 200 pieces of equipment showed that 27 of them were faulty. Test their claim at the 5% significance level?arrow_forwardA national organization that conducts research on the cost and quality of health care in the U.S. reported that, in 2012, U.S. families spent an average of $9,590 on health care expenses. Suppose you decide to test whetherthe average in 2015 is greater than the average in 2012. After conducting the appropriate statistical test, you find a P-value of 0.022. If the level of significance is 0.05, which of the following is the best interpretation of the P-value? a. The P-value of 0.022 indicates that there is a 2.2% chance that the 2015 average is greater than the average amount spent in 2012.b. The P-value of 0.022 provides weak evidence that the average in 2015 average is greater than the average amount spent in 2012.c. The P-value of 0.022 provides strong evidence that the 2015 average is greater than the average amount spent in 2012.d. The P-value of 0.022 indicates that there is a very low probability that the 2015 average is different than average amount spent in 2012.arrow_forwardIn a previous poll, 38% of adults with children under the age of 18 reported that their family ate dinner together seven nights. suppose that, in a more recent poll, 399 of 1099 adults with children under the age of 18 reported that their family ate dinner together seven nights a week. is there sufficient evidence that the proportion of families with children under the age of 18 who eat diner together seven nights a week has decreased? use the a=0.05 significance levelarrow_forward
- Suppose IQ scores were obtained for 20 randomly selected sets of siblings. The 20 pairs of measurements yield x = 102.13, y = 101.35, r= 0.927, P-value = 0.000, and y = 2.05 + 0.97x, where x represents the IQ score of the older child. Find the best predicted value of y given that the older child has an IQ of 103? Use a significance level of 0.05. Click the icon to view the critical values of the Pearson correlation coefficient r. Critical Values of the Pearson Correlation Coefficient r The best predicted value of y is (Round to two decimal places as needed.) Critical Values of the Pearson Correlation Coefficient r a = 0.05 a = 0.01 NOTE: To test Ho: p=0 Jagainst H,: p#0, reject H, lif the absolute value of r is greater than the critical value in the table. 4. 0.950 0.990 0.878 0.959 6 0.811 0.917 0.754 0.875 8. 0.707 0.834 0.666 0.632 0.602 9 0.798 10 0.765 11 0.735 12 0.576 0.708 13 0.553 0.684 14 0.532 0.661 15 0.514 0.641 16 0.497 0.623 17 0.482 0.606 0.468 0.456 18 0.590 19 0.575…arrow_forwardThe US Department of Energy reported that 46% of homes were heated by natural gas. A random sample of 325 homes in Oregon found that 126 were heated by natural gas. Test the claim that proportion of homes in Oregon that were heated by natural gas is different than what was reported. Use a 1% significance level. Give answer to at least 4 decimal places. What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.) Ho: Select an answerv H4: Select an answer v? v Based on the hypotheses, compute the following: Test Statistic = p-value = Based on the above we choose to Select an answer The correct summary would be: Select an answer homes in Oregon that were heated by natural gas is different than what the DOE reported value of 46%. that the proportion ofarrow_forwardIn a previous poll, 37% of adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Suppose that, in a more recent poll, 396 of 1132 adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Is there sufficient evidence that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has decreased? Use the a= 0.05 significance level. Because npo (1-pPo) = 10 and the sample size is (Round to one decimal place as needed.) 5% of the population size, and the sample V the requirements for testing the hypothesis V satisfied What are the null and alternative hypotheses? V versus H, O Ho (Type integers or decimals. Do not round.) Find the test statistic, Zo Z0= (Round to two decimal places as needed.) Find the P-value P-value = (Round to three decimal places as needed.) Is there sufficient evidence that the proportion of families with children…arrow_forward
arrow_back_ios
arrow_forward_ios
Recommended textbooks for you
- A First Course in Probability (10th Edition)ProbabilityISBN:9780134753119Author:Sheldon RossPublisher:PEARSON
A First Course in Probability (10th Edition)
Probability
ISBN:9780134753119
Author:Sheldon Ross
Publisher:PEARSON