Question
1.Medical Specialities The following doctors were observed on staff at a local hospital. MD Doctor of Osteopathy Pathology 8 1 Pediatrics 5 2 Orthopedics 12
1.Medical Specialities The following doctors were observed on staff at a local hospital.
MD Doctor of Osteopathy
Pathology 8 1
Pediatrics 5 2
Orthopedics 12 2
Choose one doctor at random. Find the probabilities of the following. Express your answer as a fraction or a decimal rounded to three decimal places.Part 1 of 2
What is the probability that she is a pediatrician?
P(pediatrician)= ?
Part 2 of 2
What is the probability that he is an orthopedist or a Doctor of Osteopathy?
P(orthopedist or Doctor of Osteopathy) = ?
2.Gift Baskets The Gift Basket Store has the following premade gift baskets containing the following combinations in stock.
Cookies Mugs Candy
Coffee 16 17 18
Tea 15 18 15
Choose 1 basket at random. Find the probability that it contains the following combinations.
Enter your answers as fractions or as decimals rounded to 3 decimal places.Part 1 of 3
(a) Tea or cookies
P(tea or cookies)=?
Part 2 of 3
(b) Coffee, given that it contains candy
P(coffee, given that it contains candy)=?
Part 3 of 3
(c) Coffee and mugs
P(coffee and mugs)=?
3.Guidance Missile System A missile guidance system has seven fail-safe components. The probability of each failing is 0.1. Assume the variable is binomial. Find the following probabilities. Do not round intermediate values. Round the final answer to three decimal places.
(a)Exactly three will fail.
P(exactly three will fail) = ?
(b)More than three will fail.
P(more than three will fail) = ?
(c)All will fail.
P(all will fail) = ?
(d)Compare the answers for parts a, b, and c, and explain why these results are reasonable.
Since the probability of each event becomes less likely, the probabilities become
4.Monthly Mortgage Payments The average monthly mortgage payment including principal and interest is $982 in the United States. If the standard deviation is approximately $180 and the mortgage payments are approximately normally distributed, find the probabilities. Use a TI-83 Plus/TI-84 Plus calculator and round the answers to at least four decimal places.
(a)The selected monthly payment is more than $1200.
P(X>1200) = ?
(b)The selected monthly payment is between $750 and $1050.
P(750 5.The average number of moves a person makes in his or her lifetime is 12 and the standard deviation is 3.3. Assume that the sample is taken from a large population and the correction factor can be ignored. Round the final answers to four decimal places and intermediate z value calculations to two decimal places. Part 1 of 3 Find the probability that the mean of a sample of 25 people is less than 10. P(X<10)=? Part 2 of 3 Find the probability that the mean of a sample of 25 people is greater than 10. P(X>10)=? Part 3 of 3 Find the probability that the mean of a sample of 25 people is between 11 and 12. P(11 6.The numbers of faculty at 20 randomly selected state-controlled colleges and universities with enrollment under 12,000 students are shown below. Use these data to estimate the mean number of faculty at all state-controlled colleges and universities with enrollment under 12,000 with 98% confidence. Assume =165.1. Round intermediate and final answers to one decimal place. Assume the population is normally distributed. 515 280 289 180 431 176 318 836 203 374 412 134 539 471 211 384 396 224 337 395 < < 7.State Gasoline Taxes A random sample of state gasoline taxes (in cents) is shown here for 9 states. Round sample statistics and final answers to at least two decimal places. 48.9 42.1 64.2 43.6 43.5 42.1 55.5 47.2 39.2 Part 1 of 2 Use the data to estimate the true population mean gasoline tax with 90% confidence. Assume the variable is normally distributed. < < Part 2 of 2 Does the interval contain the national average of 44.7 cents? (Choose one), the national average of 44.7 cents (Choose one) in the interval 8.Cost of Braces The average cost for teeth straightening with metal braces is approximately $5481. A nationwide franchise thinks that its cost is below that figure. A random sample of 24 patients across the country had an average cost of $5152 with a standard deviation of $624. At =0.10, can it be concluded that the mean is less than $5481? Assume that the population is approximately normally distributed. (a) State the hypotheses and identify the claim. :H0 (Choose one) $ (Choose one) :H1 (Choose one) $ (Choose one) This hypothesis test is a (Choose one) test. (b) Find the critical value(s). Round the answer(s) to at least three decimal places. If there is more than one critical value, separate them with commas. Critical value(s): ? Part 3 of 5 (c) Compute the test value. Always round t score values to at least three decimal places. t=? Part 4 of 5 (d) Make the decision. (Choose one) the null hypothesis 9.Life on Other Planets Forty-six percent of people believe that there is life on other planets in the universe. A scientist does not agree with this finding. He surveyed 125 randomly selected individuals and found 75 believed that there is life on other planets. At =0.05, is there sufficient evidence to conclude that the percentage differs from 46%? Use the critical value method with tables. Part 1 of 5 (a)State the hypotheses and identify the claim. :H0 (Choose one) :H1 (Choose one) This hypothesis test is a (Choose one) test. Part 2 of 5 (b)Find the critical value(s). Round the answer to two decimal places. If there is more than one critical value, separate them with commas. Critical value(s): ? Part 3 of 5 (c)Compute the test value. Round the answer to at least two decimal places. z = ? Part 4 of 5 (d)Make the decision. (Choose one) the null hypothesis 10.Telephone Calls A researcher knew that before cell phones, a person made on average 2.5 calls per day. He believes that the number of calls made per day today is higher. He selects a random sample of 34 individuals who use a cell phone and asks them to keep track of the number of calls that they made on a certain day. The mean was 2.6. At =0.01, is there enough evidence to support the researcher's claim? The standard deviation for the population found by a previous study is 0.6. Assume that the variable is normally distributed. Use the P-value method with tables. State the hypotheses and identify the claim. :H0 (Choose one) :H1 (Choose one) The hypothesis test is a (Choose one) test. Part 2 of 5 Compute the test value. Always round z score values to at least two decimal places. z=? Part 3 of 5 Find the P-value. Round the answer to at least four decimal places. P-value=? Part 4 of 5 Make the decision. (Choose one) the null hypothesis. 11.Reject or Not State whether the null hypothesis should be rejected on the basis of the given P-value: P-value=0.0160, =0.05, one-tailed test Reject the null hypothesis. Do not reject the null hypothesis. 12.Educational Television In a random sample of 200 people, 154 said that they watched educational television. Find the 90% confidence interval of the true proportion of people who watched educational television. Round intermediate answers to at least five decimal places. Round your final answers to at least three decimal places. < p < `13.A researcher wishes to see if there is a difference between the mean number of hours per week that a family with no children participates in recreational activities and a family with children participates in recreational activities. She selects two random samples and the data are shown. Use 1 for the mean number of families with no children. At =0.10, is there a difference between the means? Use the P-value method and tables. X n No children 8.9 2.9 20 Children 10.1 2.9 20 Part 1 of 5 State the hypotheses and identify the claim with the correct hypothesis. H0: (Choose one) H1: (Choose one) This hypothesis test is a Part 2 of 5 Compute the test value. Round all intermediate steps to four decimal places and always round z score values to two decimal places. z=? Part 3 of 5 Compute the P-value. Round the answer to four decimal places. P-value =? Part 4 of 5 Make the decision. (Choose one) the null hypothesis. 14.Waterfall Heights Is there a significant difference at =0.01 in the mean heights in feet of waterfalls in Europe and the ones in Asia? The data are shown. Use the critical value method with tables. Europe Asia 487 1312 820 614 722 470 345 350 964 900 1385 722 830 Send data to Excel Use 1 for the mean height of waterfalls in Europe. Assume the variables are normally distributed and the variances are unequal. Part 1 of 5 State the hypotheses and identify the claim with the correct hypothesis. H0 : (Choose one) H1 : (Choose one) This hypothesis test is a (Choose one) test Part 2 of 5 Find the critical value(s). Round the answer(s) to three decimal places. If there is more than one critical value, separate them with commas. Critical value(s): ? Part 3 of 5 Compute the test value. Always round t score values to three decimal places. t=? Part 4 of 5 Make the decision. (Choose one) the null hypothesis. 15.Victims of Violence A random survey of 90 women who were victims of violence found that 24 were attacked by relatives. A random survey of 60 men found that 10 were attacked by relatives. At =0.05, can it be shown that the percentage of women who were attacked by relatives is different from the percentage of men who were attacked by relatives? Use p1 for the proportion of women who were attacked by relatives. Use the P-value method with tables Part 1 of 5 State the hypotheses and identify the claim with the correct hypothesis. H0 : (Choose one) H1 : (Choose one) This hypothesis test is a (Choose one) test. Part 2 of 5 Compute the test value. Round sample proportions to three decimal places and z score values to two decimal places. z-value=? Part 3 of 5 Find the P-value. Round the answer to four decimal places. P-value=? Part 4 of 5 Make the decision. (Choose one) the null hypothesis.
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