Two treatments for a new infectious disease have been developed to reduce hospitalizations. Each treatment was given to different groups of patients. Treatment 1 was given to 65 people. Subjects of treatment 1 had a
(a) List the null and alternative hypotheses for this test.
(b) Compute the value of the test statistic
(c) Determine the rejection region.
(d) Determine whether or not we can reject the null hypothesis.
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- The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maximum possible rating is 10. Suppose a simple random sample of 50 business travelers is selected and each traveler is asked to provide a rating for the Miami International Airport. The ratings obtained from the sample of 50 business travelers provides a mean of 6.34 and a standard deviation of 2.163. Develop a 95% confidence interval estimate of the population mean rating for Miami.arrow_forwardIn a random sample of 17 people, the mean commute time to work was 30.2 minutes and the standard deviation was 7.1 minutes. Assume the population is normally distributed and use a t- distribution to construct a 95% confidence interval for the population mean y. What is the margin of error of y? Interpret the results.arrow_forwardA custodian wishes to compare two competing floor waxes to decide which one is best. He believes that the mean of WaxWin is not equal to the mean of WaxCo. In a random sample of 12 floors of WaxWin and 20 of WaxCo. WaxWin had a mean lifetime of 27.6 with a standard deviation of 8.8 and WaxCo had a mean lifetime of 20.3 with a standard deviation of 7.6. Perform a hypothesis test using a significance level of 0.01 to help him decide. Let WaxWin be sample 1 and WaxCo be sample 2 The correct hypotheses are: HA: P1 > µ2(claim) Ho: µ1 2 µ2 HA: µ1 < µ2(claim) Ho: µ1 = µ2 HA: H1 + H2(claim) Since the level of significance is 0.01 the critical value is 2.836 and -2.836 The test statistic is: (round to 3 places) The p-value is: (round to 3 places)arrow_forward
- A study was conducted to estimate hospital costs for car accident victims who woreseat belts. Suppose that 46 randomly selected cases had an average of $15150and a standard deviation of $8500. If the average hospital costs for those that didnot wear seatbelts was $19000, construct a 90% confidence interval to see if theaverage costs are significantly different.arrow_forwardThe mean age of bus drivers in chicago is 56.7 years. If a hypothesis test is performed, how should you interpret a decision that rejects the null hypothesis?arrow_forwardOI al College is less likely to attend orientation than men before they begin their coursework. A random sample of freshmen at Oxnard College were asked what their gender is and whether they attended orientation. The results of the survey are shown below: Data for Gender vs. Orientation Attendance Select an answer Women Men 439 242 What can be concluded at the a = 0.01 level of significance? For this study, we should use Select an answer a. The null and alternative hypotheses would be: Yes 350 No 262 Ho: Select an answer ✓ Select an answer ✓ Select an answer ✓ (please enter a decimal and note that p1 and μl represent the proportion and mean for women and p2 and μ2 represent the proportion and mean for men.) H₁: b. The test statistic ? | = c. The p-value d. The p-value is ? ✓ a e. Based on this, we should f. Thus, the final conclusion is that ... Select an answer ✓ Select an answer (Please enter a decimal) (please show your answer to 3 decimal places.) (Please show your answer to 4…arrow_forward
- A custodian wishes to compare two competing floor waxes to decide which one is best. He believes that the mean of WaxWin is less than the mean of WaxCo. In a random sample of 20 floors of WaxWin and 17 of WaxCo. WaxWin had a mean lifetime of 26.4 with a standard deviation of 8.6 and WaxCo had a mean lifetime of 27.2 with a standard deviation of 6.6. Perform a hypothesis test using a significance level of 0.05 to help him decide. Let WaxWin be sample 1 and WaxCo be sample 2 The correct hypotheses are: Ο Ηο: μι < με HA₁₂(claim) Ο Ηρ: με 2 με HA: 1₂(claim) O Ho: ₁ = ₂ HAM12(claim) Since the level of significance is 0.05 the critical value is -1.69 (round to 3 places) The test statistic is: The p-value is: (round to 3 places)arrow_forwardA technician compares repair costs for two types of microwave ovens (type I and type II). He believes that the repair cost for type I ovens is greater than the repair cost for type II ovens. A sample of 35 type I ovens has a mean repair cost of $80.39. The population standard deviation for the repair of type I ovens is known to be $24.63. A sample of 31 type II ovens has a mean repair cost of $73.47. The population standard deviation for the repair of type II ovens is known to be $10.42. Conduct a hypothesis test of the technician's claim at the 0.05 level of significance. Let i be the true mean repair cost for type I ovens and µz be the true mean repair cost for type II ovens. Step 1 of 5: State the null and alternative hypotheses for the test.arrow_forwardJess2.4 employees leaving per month, with a standard deviation of 0.5. She decided to see if using a new benefits package would affect her average. She used the new benefits package for 13 months. For those 13 months, Jessica got an average of 2.6 employees leaving. She wants to know if the new benefits package caused her to get a different number of employees leaving per month. She conducts a one-mean hypothesis at the 5% significance level, to see if the new benefits package caused her to get a different number of employees leaving per month. Which answer choice shows the correct null and alternative hypotheses for this test? Select the correct answer below: H0:μ=2.4; Ha:μ≠2.4, which is a two-tailed test. H0:μ=2.4; Ha:μ<2.4, which is a left-tailed test. H0:μ=2.4; Ha:μ>2.4, which is a right-tailed test. H0:μ=2.6; Ha:μ≠2.6, which is a left-tailed test.arrow_forward
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