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- a) Soil has been improved to increase the strength of a ground. The average compressive strength of 61 samples before treatment is 45 kg / cm2 and the standard deviation is 6.75 kg / cm2; The average compressive strength of 56 samples after the improvement is 52.5 kg / cm2 and the standard deviation is 8 kg / cm2. Determine whether the improvement application changes the standard deviation of the soil strength or not at the 5% significance level .b) Find the confidence intervals at the 7% significance level of the mean compressive strength of the sample after improvement?Bone mineral density (BMD) is a measure of bone strength. Studies show that BMD declines after age 45. The impact of exercise may increase BMD. A random sample of 59 women between the ages of 41 and 45 with no major health problems were studied. The women were classified into one of two groups based upon their level of exercise activity: walking women and sedentary women. The 39 women who walked regularly had a mean BMD of 5.96 with a standard deviation of 1.22. The 20 women who are sedentary had a mean BMD of 4.41 with a standard deviation of 1.02. Which of the following inference procedures could be used to estimate the difference in the mean BMD for these two types of womenDo men talk less than women? The table shows results from a study of the words spoken in a day by men and women. Assume that the two samples are randomly selected, independent, the population standard deviations are not know and not considered equal. At the 0.05 significance level, test the claim that the mean number of words spoken by men is less than the mean number of words spoken by women. Men Women n1 = 212 n2 = 218 xˉx̄1 = 15706.4 words xˉx̄2 = 15683.2 words s1 = 1595.44 words s2 = 1597.54 words What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.)H0: Select an answer μ₁ μ σ₁² x̄₂ p x̄₁ s₁² p̂₁ μ(men) p₂ p₁ μ₂ ? ≤ > ≥ = ≠ < Select an answer μ₁ p₁ μ₂ x̄₂ σ₁² p̂₁ μ s₁² μ(women) p x̄₁ p₂ H1: Select an answer p̂₂ p s₂² σ₂² μ(men) μ μ₂ x̄₂ x̄₁ p₂ μ₁ p₁ ? < > = ≠ ≥ ≤ Select an answer μ₂ μ p₂ s₁² p₁ x̄₁ σ₁² p̂₁ p μ₁ x̄₂ μ(women) Original Claim = Select an answer H₁ H₀ df = Based on the hypotheses, find the…
- In tests of a computer component, it is found that the mean time between failures is 909 hours. A modification is made which is supposed to increase reliability by increasing the time between failures. Tests on a sample of 25 modified components produce a mean time between failures of 955 hours. Using a 1% level of significance, perform a hypothesis test to determine whether the mean time between failures for the modified components is greater than 909 hours. Assume that the population standard deviation is 53 hours.A researcher is studying how much electricity (in kilowatt hours) people from two different cities use in their homes. Random samples of 13 days from Nashville (Group 1) and 14 days from Cincinnati (Group 2) are shown below. Test the claim that the mean number of kilowatt hours in Nashville is less than the mean number of kilowatt hours in Cincinnati. Use a significance level of a = 0.10. Assume the populations are approximately normally distributed with unequal variances. Round answers to 4 decimal places. Nashville Cincinnati 902.6 892.9 904.9 897.5 905.2 904.1 911.4 883.9 898.7 887.7 889.7 907.2 899.1 Ho: M₁ What are the correct hypotheses? Note this may view better in full screen mode. Select the correct symbols for each of the 6 spaces. 921.2 931.1 908.3 939.6 898 934.6 906.6 921.2 907.5 902.4 883.6 893.1 Test Statistic = 941.4 923.6 p-value = H₁: M₁ Based on the hypotheses, find the following: H₁₂ H₂ OrAn engineer designed a valve that will regulate water pressure on an automobile engine. The engineer designed the valve such that it would produce a mean pressure of 5.5 pounds/square inch. It is believed that the valve performs above the specifications. The valve was tested on 24 engines and the mean pressure was 5.7 pounds/square inch with a variance of 0.49. A level of significance of 0.01 will be used. Assume the population distribution is approximately normal. Make the decision to reject or fail to reject the null hypothesis.
- A sample of 60 chewable vitamin tablets have a sample mean of 298 milligrams of vitamin C. Nutritionists want to perform a hypothesis test to determine how strong the evidence is that the mean mass of vitamin C per tablet differs from 299 milligrams. State the appropriate null and alternate hypotheses.Calculate the CV of an assay that gives results for the mean and SD of 5.0 and 0.5 mmol /L.The mean fineness of yarn is expected to be greater than the standard value of 5 units. To test this claims, the factory graded 16 specimens of the yarn and found the sample mean to be 5.9 units. Assume that the measurement of fineness is normally distributed with a variance of 4.0. We can state (exactly one option is correct) The p-value is 0.1632 and there is strong evidence that the mean exceeds by 5 units at 17% level of significance. The p-value is 0.0002 and there is strong evidence that the mean exceeds 5 units at 1% level of significance. The p-value is 0.3264 which indicates the data are consistent with a mean of 5 units at 35% level of significance.
- In an automobile repair and service industry, the average number of defective items in all the lots of automobile parts produced by a machine last week was 8. The supervisor believes that the mean number of defective parts produced by the same machine will differ in the present week. To test whether the average differs from 8, he samples 30 lots to obtain the mean number of defective parts to be 9 with a standard deviation of 3.Using the critical value approach, identify the decision at the 5% level of significance. Group of answer choices The decision is not to reject H0 because the test statistic does not fall in the rejection region (-∞,-1.960] U [1.960, ∞). The decision is to reject H0 because the test statistic falls in the rejection region (-∞,-1.645]. The decision is not to reject H0 because the test statistic does not fall in the rejection region [1.645, ∞). The decision is not to reject H0 because the test statistic does not fall in the rejection region (-∞,-1.645] U…The mean length of steel bolts from a manufacturing process is advertised as 15.1 mm. An inspector believes that this value is not correct. To investigate the inspector's claim, the lengths of a random sample of size 280 bolts are measured and found to have a sample mean of 15.6 mm. We will assume that the population of bolt lengths is normally distributed with standard deviation of 5.0 mm. Test whether the mean population length has changed at a level of significance of 10%. Show your relevant steps below. Step 1: Hypotheses: • The null hypothesis Ho is (choose one): = 5.0 u = u = 15.1 OH = 15.6 u = 280.0 • For the alternative hypothesis H. we change the equal sign in Ho to (choose one): Step 2: Rejection Region: The rejection region is bounded by the following critical values: Critical z value(s) =A researcher decides to measure anxiety in group of bullies and a group of bystanders using a 23-item, 3 point anxiety scale. Assume scores on the anxiety scales are normally distributed and the variance among the group of bullies and bystanders are the same. A group of 30 bullies scores an average of 21.5 with a sample standard deviation of 10 on the anxiety scale. A group of 27 bystanders scored an average of 25.8 with a sample standard deviation of 8 on the anxiety scale. You do not have any presupposed assumptions whether bullies or bystanders will be more anxious so you formulate the null and alternative hypothesis based on that.
