A sample of 14 joint specimens of a particular type gave a sample mean proportional limit stress of 8.48 MPa and a sample standard deviation of 0.79 MPa
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A sample of 14 joint specimens of a particular type gave a sample mean proportional limit stress of 8.48 MPa and a sample standard deviation of 0.79 MPa
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- Following maintenance and calibration, an extrusion machine produces aluminum tubing with a mean outside diameter of 2.500 inches, with a standard deviation of 0.027 As the machine functions over an extended number of work shifts, the standard deviation remains unchanged, but the combination of accumulated deposits and mechanical wear causes the mean diameter to “drift” away from the desired 2.500 inches. For a recent random sample of 34 tubes, the mean diameter was 2.509 inches. At the 0.01 level of significance, does the machine appear to be in need of maintenance and calibration?The specific gravities of a simple random sample of 91 samples of a particular wort has a mean x ̄ = 1.05 platos and a standard deviation sx = 0.16 platos. Use a 0.05 significance level to test the claim that the mean specific gravity of this wort is less than 1.10.A simple random sample of 50 men from a nomally distributed population results in a standard deviation of 9.8 beats per minute. The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.10 significance level to test the claim that pulse rates of men have a standard deviation equal to 10 beats per minute Complete parts (a) through (d) below. a. Identify the null and altemative hypotheses. Choose the corect answer below. B. Ho: 0= 10 beats per minute A Hg: 02 10 beats per minute H:o< 10 beats per minute H:0 10 beats per minute C. H:o= 10 beats per minute O D. Ho: o 10 beats per minute H:0= 10 beats per minute H o<10 beats per minute b. Compute the test statistic. (Round to three decimal places as needed.)
- Question Help v A simple random sample of pulse rates of 20 women from a normally distributed population results in a standard deviation of 10.7 beats per minute. The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.01 significance level to test the claim that pulse rates of women have a standard deviation equal to 10 beats per minute. Complete parts (a) through (d) below. a. Identify the null and alternative hypotheses. Choose the correct answer below. O A. Ho: o± 10 beats per minute O B. Ho: o = 10 beats per minute H1: o<10 beats per minute H: 0 = 10 beats per minute O C. Ho: o 10 beats per minute O D. Ho: o = 10 beats per minute %3! H,: o<10 beats per minute H: o7 10 beats per minute b. Compute the test statistic. x = D (Round to three decimal places as needed.) %3D c. Find the P-value of the…3) A study was conducted to see if increasing the substrate concentration has an appreciable effect on the velocity of a chemical reaction. With a substrate concentration of 1.5 moles per liter, the reaction was run 15 times, with an average velocity of 7.5 micromoles per 30 minutes and a standard deviation of 1.5. With a substrate concentration of 2.0 moles per liter, 12 runs were made, yielding an average velocity of 8.8 micromoles per 30 minutes and a sample standard deviation of 1.2. Is there any reason to believe that this increase in substrate concentration causes an increase in the mean velocity of the reaction of more than 0.5 micromole per 30 minutes? Use a 0.01 level of significance and assume the populations to be approximately normally distributed with equal variancesA sample of 11 joint specimens of a particular type gave a sample mean proportional limit stress of 8.55MPa and a sample standard deviation of 0.71 MPa. Calculate and interpret a 90% lower confidence bound for the true average proportional limit stress of all such joints. ( round your answer to two decimal places). MPa Interpret this bound. O with 90% confidence, we can say that the value of the true mean proportional limit stress of all such joints is greater than this value. O with 90% confidence, we can say that the value of the true mean proportional limit stress of all such joints is centered around this value. O with 90% confidence, we can say that the value of the true mean proportional limit stress of all such joints is less than this value.
- Let type A thread correspond to population 1 and let type B thread correspond to population 2. Identify the null and alternative hypotheses. Choose the correct answer below. O A. Ho H4- H2# 14 H1 H1 - H2 = 14 OD. Ho H-H2> 14 H, H - 12 = 14 B. Ho H - 42 = 14 OC. Ho H1 -42 = 14 H -2 14 OF. H H1-H2 = 14 H -12 > 14 OE Ho H-H2 14 H, -2=14 Identify the critical region. Select the correct choice below and fill in the answer box(es) to complete your choice. (Round to two decimal places as needed) XA. I' B. t'> 1.66 Oc. t'< Find the test statistic 2.73 (Round to two decimal places as needed)In a study of the invasive species heracleum mantegazzium, Jan Pergl and associates compared the density of these plants in their native area in Russia and in an invaded area of the Czech Republic. In their native area, the average density was well-established as μ = 5 plants per square meter. In an invaded area, a sample of n = 60 plants produced an average density of x = 11.117 plants per square meter with a sample standard deviation of s = 2.37 plants per square meter. Using a signficance level of α = 0.01, is there sufficient evidence to conclude that the density of the species in the invasive area is significantly higher than in the native area? Use the hypothesis-testing framework to answer this question. Be sure to state both the null and alternative hypothesis, calculate a test statistic, and state the conclusion in non-technical terms. You may use either the P-value or critical value method, but make sure to state which method you use.The slant shear test is widely accepted for evaluating the bond of resinous repair materials to concrete; it utilizes cylinder specimens made of two identical halves bonded at 30°. An article reported that for 12 specimens prepared using wire-brushing, the sample mean shear strength (N/mm2) and sample standard deviation were 18.60 and 1.56, respectively, whereas for 12 hand-chiseled specimens, the corresponding values were 23.19 and 4.07. Does the true average strength appear to be different for the two methods of surface preparation? State and test the relevant hypotheses using a significance level of 0.05. (Use ?1 for wire-brushing and ?2 for hand-chiseling.) H0: ?1 − ?2 = 0Ha: ?1 − ?2 ≤ 0H0: ?1 − ?2 = 0Ha: ?1 − ?2 < 0 H0: ?1 − ?2 = 0Ha: ?1 − ?2 > 0H0: ?1 − ?2 = 0Ha: ?1 − ?2 ≠ 0 Calculate the test statistic and determine the P-value. (Round your test statistic to one decimal place and your P-value to three decimal places.) t = P-value = You may have computed…
- A fishing line manufacturer tests the breaking stain of his products. He has found that, for a particular ‘line weight’, the maximum breaking stain follows a normal distribution, with a mean of 21.5kg and a standard deviation of 1.5kg. If a line is selected at random, calculate the probability that it has a breaking strain of between 22kg and 23.5kg.The men body mas index (BMI) for boys of age 12 years is 23.6kg/m An investigator wants to test if the BMI is higher in 12-year-old boys living in New York CIty. How many bos are needed to ensure that a two-sided test of hypothesis has 80% power to detect a difference in BMI of 2 kg/m? Assume that the standard deviation in BMI is 5.7 kg/m.A producer (spinner) supplies yarn of nominal linear density equal to be 45 tex. The customer (knitter) accepts yarn if its mean linear density lies within a range of 45±1.5 tex. As the knitter cannot test all the yarns supplied by the spinner, the knitter would like to devise an acceptance sampling scheme with 10% producer's risk and 5% consumer's risk. Assume the standard deviation of count within a delivery is 1.2 tex.