A civil engineer is analysing the compressive strength of concrete. A random sample of 12 specimens has a mean compressive strength of 3201.33 psi and a standard deviation of 900 psi. Construct a 95% confidence interval for the mean compressive strength. 4.
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- A local bottler in Hawaii wishes to ensure that an average of 10 ounces of passion fruit juice is used to fill each bottle. In order to analyze the accuracy of the bottling process, he takes a random sample of 39 bottles. The mean weight of the passion fruit juice in the sample is 9.54 ounces. Assume that the population standard deviation is 1.25 ounce. (You may find it useful to reference the appropriate table: z table or t table) Select the null and the alternative hypotheses to test if the bottling process is inaccurate. multiple choice 1 H0: μ = 10; HA: μ ≠ 10 H0: μ ≤ 10; HA: μ > 10 H0: μ ≥ 10; HA: μ < 10 b-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) b-2. Find the p-value. 0.05 p-value < 0.10 p-value 0.10 p-value < 0.01 0.01 p-value < 0.025 0.025 p-value < 0.05 c-1. What is the conclusion at α = 0.01?…The fill of 100 gram paxkets of peanuts is normally distributed with a mean of 99.8 grams and standard deviation of 0.15 grams. To what mist the standard deviation of the packets be adjusted if only 5% of the paxkets are to be overfilled.Can someone please tell me why my t value is wrong? ASAP??!!!
- Please label each part.A company that makes cola drinks states that the mean caffeine content per 12-ounce bottle of cola is 55 milligrams. You want to test this claim. During your tests, you find that a random sample of thirty 12-ounce bottles cola has a mean caffeine content of 57.4 milligrams. Assume the population is normally distributed and the population standard deviation is 6.9 milligrams. At a = 0.04, can you reject the company's claim? Complete parts (a) through (e). C (a) Identify Ho and H₂. Choose the correct answer below. OA. Ho: 257.4 OB. Ho: H=57.4 H_:μ#57.4 H₂H<57.4 C. Ho: H=55 H₂:μ#55 OD. Ho: μ255 H:H<55 OE. Ho: 57.4 OF. Ho: *55 H₂:μ=57.4 Ημ= 55 (b) Find the critical value(s). Select the correct choice below and fill in the answer box within your choice. (Round to two decimal places as needed.) A. The critical values are ± 2.05. OB. The critical value is. Identify the rejection region(s). Choose the correct answer below. OA OB. O C. O Fail to reject Ho Fail to reject Ho Fail to reject Ho Q X…A company's single-serving cereal boxes advertise 1.63 ounces of cereal. In fact, the amount of cereal X in a randomly selected box can be modeled by a Normal distribution with a mean of 1.70 ounces and a standard deviation of 0.03 ounces. Let Y = the excess amount of cereal beyond what's advertised in a randomly selected box, measured in grams (1 ounce = 28.35 grams). A. Find the mean of Y B. Calculate and interpret the standard deviation of Y C. Find the probability of getting at least 1 gram more cereal than advertised
- A cement manufacturer claims that the average weight of his product is 40 kgs. A sample of 45 bags was taken and found to have a mean weight of 38.71 kgs. with a standard deviation of 2.35 kgs. Are we going to accept the claim of the cement manufacturer? Use α = 0.01."Durable press" cotton fabrics are treated to improve their recovery from wrinkles after washing. "Wrinkle recovery angle" measures how well a fabric recovers from wrinkles. Higher is better. Here are data on the wrinkle recovery angle (in degrees) for two types of treated fabrics: Permafresh Hylite 15 17 12 14 13 16 16 15 14 18 A manufacturer wants to know how large is the difference in mean wrinkle recovery angle. Give a 98% confidence interval for the difference in mean wrinkle recovery angle: [three decimal accuracy] [three decimal accuracy]In the 1800s, German physician Carl Reinhold, took millions of axillary (i.e. armpit) temperatures from soldiers. This study established that body temperature is normally distributed and the standard normal human body temperature is 98.6°F with a standard deviation of 0.72 °F. In a recent study, American researchers obtained 5,000 axillary temperatures from a Los Angeles hospital. The mean of these temperature readings was 97.9 F. Assuming a Type I error risk of no more than 5%, did the findings support the theory that human, body temperature has decreased since the 1800s? What is the Zcrit?
- It takes an average of 10.4 minutes for blood to begin clotting after an injury. An EMT wants to see if the average will increase if the patient is immediately told the truth about the injury. The EMT randomly selected 67 injured patients to immediately tell the truth about the injury and noticed that they averaged 10.8 minutes for their blood to begin clotting after their injury. Their standard deviation was 2.94 minutes. What can be concluded at the the a = 0.05 level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: 10.4 H1: 1.5 c. The test statistic t 1.114 (please show your answer to 3 decimal places.) d. The p-value = 0.0500 (Please show your answer to 4 decimal places.) e. The p-value is ? va f. Based on this, we should accept g. Thus, the final conclusion is that ... v the null hypothesis. O The data suggest that the population mean is not significantly greater than 10.4 at a = 0.05, so there is statistically…The average IQ level in the United States is 98 (standard deviation of 4). A sample of 200 people was selected from the city of Loma Linda and had an IQ average of 105. Assuming an alpha of 0.05, is the sample mean significantly greater than the population mean? 1. List the 5 steps needed for the hypothesis testing. Do not forget to make your conclusion in step 5. 2. Compute and interpret the effect size. Edit View Insert Format Tools Table 12pt ✓ Paragraph BIUA ✓ T² vA gallon of paint claims to cover 400 sq ft. This feels too high. To test this, a research lab obtained 30 random gallons of paint and recruited 30 volunteers to each paint as much wall space as they could with a gallon, and the lab techs measured the area covered. They computed a mean of 396 sq ft with a standard deviation of 27 sq ft. They performed a test of hypothesis and computed a p-value of 0.212. What is the appropriate conclusion?