An agronomist believes that a newly developed fertilizer will increase the mean harvest of eggplants by more than 2.5 kg. 56 plats were treated with fertilizer and had a mean harvest of 10.5 kg with standard deviation of 1.2 kg. 29 plants were untreated and had a mean harvest of 9.25 kg with standard deviation of 0.7 kg. What is the null hypothesis of the problem? p> 2.5 kg Ob. p= 9.25 9.25< 10.5 kg Od. H= 10.5
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- An electronics production sequence includes a process where a film of metal is deposited on a board. The film must be more than 1.000 mm thick. If the film is too thin the process fails. The engineer ran a test where the null hypothesis was that the mean thickness was at most 1.000 mm. A sample of 100 boards had a mean thickness of 1.007 mm and a standard deviation of 0.050 mm.At what level of significance would the null hypothesis be rejected? Express your answer at a decimal (not a percent) rounded to three places.(b) Perform a Z-test and find the p-value. Here is some information to help you with your Z-test. • The value of the test statistic is given by √n The p-value is the area under the curve to the right of the value of the test statistic. O One-tailed o Two-tailed Step 2: Enter the test statistic. (Round to 3 decimal places.) Step 3: Shade the area represented by the p- value. Step 4: Enter the p- value. (Round to 3 decimal places.) 0.3 0.2 0.1 (c) Based on your answer to part (b), choose what can be concluded, at the 0.05 level of significance, about the claim made by the studio. o Since the p-value is less than (or equal to) the level of significance, the null hypothesis is rejected. So, there is enough evidence to support the claim that the mean player rating is higher than 90. Since the p-value is less than (or equal to) the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to support the claim that the mean player rating is higher than 90.…A leasing firm claims that the mean number of miles driven annually, µ, in its leased cans is less than 12460 miles. A random sample of 27 cars leased from this firm had a mean of 11839 annual miles driven. It is known that the population standard deviation of the number of miles driven in cars from this firm is 1620 miles. Assume that the population is normally distributed. Is there support for the firm’s claim at the 0.05 level of significance? The null hypothesis: H0 : _____ The alternative hypothesis: H1: ______ The type of test statistic: (choose one) _______ The value of the test statistic: ______ (round to at least three decimal places.) The p-value: _____ (round to at least three decimal places.) Can we support the leasing firm’s claim that the mean number of miles driven annually is less than 12460 miles? Yes _____, NO ______
- 2. A pain reliever currently being used in a hospital is known to bring relief in a mean time of 3.5 minutes. To compare a new pain reliever with the one currently being used, a random sample of 45 patients is given the new drug. The mean time to relief for this sample is 2.8 minutes with a standard deviation of 1.14 minutes. a) Test the research (alternative) hypothesis that the new drug is more effective than the old drug, using α ≤ .05. b) Assuming a standard deviation of 1.14 minutes, approximately how large a sample of patients would be needed to form a 90% confidence interval to estimate the population time-to-relief of the new drug accurate to within plus or minus 10 seconds? Answers: 2. a) Z = -4.12, reject H0, new drug is more effective, b) 1276. An automatic Coffee dispenser is advertised to pour u=16 oz of coffee with a standard deviation of o-0.1 oz. An unsatisfied customer wants to know if the automatic Coffee dispenser pours less than 16 oz on average. A random sample of n=49 cups of coffee is taken and is found to have a sample mean x = 15.96 oz. Use a a-5% level of significance to test if the population mean u is less than 16 oz. a) State the null and alternate hypothesis. b) Find the test statistic (either z or t). c) Find the p-value or sketch the critical region. d) State whether you will reject or fail to reject the nullA carpenter is making doors that are 2058.0 millimeters tall. If the doors are too long they must be trimmed, and if they are too short they cannot be used. A sample of 17 doors is made, and it is found that they have a mean of 2045.0 millimeters with a standard deviation of 5.0. Is there evidence at the 0.05 level that the doors are too short and unusable? Assume the population distribution is approximately normal. Step 1 of 3: State the null and alternative hypotheses. Step 2: Find the P-value for the hypothesis test. Round your answer to four decimal places. Step 3:
- A coin-operated drink machine was designed to discharge a mean of 8 ounces of coffee per cup. In a test of the machine, the discharge amounts in 20 randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were 8.14 ounces and 0.23 ounces, respectively. If we assume that the discharge amounts are normally distributed, is there enough evidence, at the 0.05 level of significance, to conclude that the true mean discharge, μ , differs from 8 ounces? Perform a two-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. The null hypothesis: H0: The alternative hypothesis: H1: The type of test statistic: (Choose one)ZtChi squareF The value of the test statistic:(Round to at least three decimal places.) The two critical values at the 0.05 level…Suppose the mean wait-time for a telephone reservation agent at a large airline is 44 seconds. A manager with the airline is concerned that business may be lost due to customers having to wait too long for an agent. To address this concern, the manager develops new airline reservation policies that are intended to reduce the amount of time an agent needs to spend with each customer. A random sample of 250 customers results in a sample mean wait-time of 43.3 seconds with a standard deviation of 4.3 seconds. Using a = 0.05 level of significance, do you believe the new policies were effective in reducing wait time? Do you think the results have any practical significance? Determine the null and alternative hypotheses. Ho: ▼44 seconds 44 seconds H₁: Calculate the test statistic. ▼ (Round to two decimal places as needed.) Calculate the P-value. P-value= (Round to three decimal places as needed.) State the conclusion for the test. O A. Reject H, because the P-value is less than the α = 0.05…A coin- ed drink machine was designed to discharge a mean of 7 ounces of coffee per cup. In a test of the machine, the discharge amounts in 13 randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were 6.96 ounces and 0.14 ounces, respectively. If we assume that the discharge amounts are normally distributed, is there enough evidence, at the 0.1 level of significance, to conclude that the true mean discharge, u, differs from 7 ounces? Perform a two-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.) (a) State the null hypothesis H and the alternative hypothesis H,. p H, :0 H :0 (b) Determine the type of test statistic to use. (Choose one) ▼ D=0 OSO (c) Find the value of the test statistic. (Round to three or more decimal places.) OSuppose the mean wait-time for a telephone reservation agent at a large airline is 42 seconds. A manager with the airline is concerned that business may be lost due to customers having to wait too long for an agent. To address this concern, the manager develops new airline reservation policies that are intended to reduce the amount of time an agent needs to spend with each customer. A random sample of 250 customers results in a sample mean wait-time of 41.3 seconds with a standard deviation of 4.2 seconds. Using a = 0.05 level of significance, do you believe the new policies were effective in reducing wait time? Do you think the results have any practical significance? www Determine the null and alternative hypotheses. Ho: P = 42 seconds. H₁: H <42 seconds Calculate the test statistic. = (Round to two decimal places as needed.) Check answer Get more help. Clear all (3) a 1:44 AM 6/19/2022 O pr BSC 7 Help me solve this View an example H Type here to search " "? * 3 Z @ 12 S E C 4 R F %…The 2013 general Social Survey asked a large number of people how much time they spent watching TV each day. The mean number of hours was 3.7 with a standard deviation of 2.5. Assume that in a sample of 57 teenagers, the sample standard deviation of daily TV time is 2.2 hours, and that the population of TV watching times is normally distributed. Under 1% significance level can you conclude that the population standard deviation of TV watching times for teenagers is different from 2.5? Procedure: One variance x² Hypothesis Test v Assumptions: (select everything that applies) ✔Population standard deviation is unknown ✔Simple random sample Normal population Sample size is greater than 30 0° The number of positive and negative responses are both greater than 10 Population standard deviation is known Step 1. Hypotheses Set-Up: Ho: Select an answer Ha: Select an answer ✓ ? V Step 2. The significance level a = OT % where? is the Select an answer ? " Part 2 of 5 and the units are and the test…Suppose the mean wait-time for a telephone reservation agent at a large airline is 40 seconds. A manager with the airline is concerned that business may be lost due to customers having to wait too long for an agent. To address this concern, the manager develops new airline reservation policies that are intended to reduce the amount of time an agont needs to spend with each customer. A random sample of 250 customers results in a sample mean wait-time of 39.1 seconds with a standard deviation of 4.2 seconds. Using a=0.05 level of significance, do you believe the new policies were effective in reducing wait time? Do you think the results have any practical significance? 6-7 (Round to two decimal places as needed.) Calculate the P-value. P.value = Round to three decimal olaces as needed.) ick to select your answer(s).