3. An experiment was performed to compare the abrasive wear of two different laminated materials. Twelve pieces of material 1 were tested by exposing each piece to a machine measuring wear. Ten pieces of material 2 were similarly tested. In each case, the depth of wear was observed. The samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4, while the samples of material 2 gave an average of 81 with a sample standard deviation of 5. Can we conclude at the 0.10 level of significance that the two unknown population variances are equal? Assume the populations to be approximately normal.

The braking ability was compared for two car models. Random samples of two cars were selected. The first random sample of size 64 cars yield a mean of 36 and a standard deviation of 8. The second sample of size 64 yield a sample mean of 33 and a standard deviation of 8. Do the data provide sufficient evidence to indicate a difference between the mean stopping distances for the two models? Use Alpha= 0.01. Ho: µ1 – µ2 = 0 vs. Ha: µ1 – µ2 + 0 .p-value = 0.0017. Reject Ho Но: Д, — м2 — 0 vs. Ha:M1 — M2 + 0. p-value — 0.034. Ассеpt Ho O Ho:µ1 – Hz Но: И — 2 — 0 vs. Ha:M, — нz + 0. p-value —D 0.0017. Ассept Ho Ho: H1 – U2 = 0 vs. Ha: µ1 – µ2 + 0. p-value = 0.034. Reject Ho

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…

Only need help with finding tcalc.

2. Traces of toxic, man-made hexachlorohexanes in North Sea sediments were extracted by a known process and by two new procedures, and measured by chromatography. Method Concentration found Conventional Procedure A (pg/g) 34.4 42.9 Standard deviation (pg/g) 3.6 1.2 Number of replications 6 6 a.) Is the standard deviation for procedure A significantly different from that of the conventional procedure at 95% confidence level? b.) Is the mean concentration found by procedure A significantly different from that of the conventional procedure at 95% confidence level?

A scientist interested in monitoring chemical contaminants in food, and thereby the accumulation of contaminants in human diets, selected a random sample of 25 male adults. From this sample, it was found that the average daily intake of dairy products was 756 grams per day with a standard deviation of 35 grams per day. Use this sample information to construct a 95% confide

A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for SUVs equipped with the tires. The mean braking distance for SUVs equipped with tires made with compound 1 is 65 feet, with a population standard deviation of 13.6. The mean braking distance for SUVs equipped with tires made with compound 2 is 69 feet, with a population standard deviation of 8.5. Suppose that a sample of 55 braking tests are performed for each compound. Using these results, test the claim that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. Let μ₁ be the true mean braking distance corresponding to compound 1 and μ₂ be the true mean braking distance corresponding to compound 2. Use the 0.05 level of significance. Step 5 of 5: State the conclusion of the hypothesis test. Answer Tables Keypad Keyboard Shortcuts Previous Step Answers There is sufficient evidence…