EXERCISE 1. A random sample of six steel beams has mean compressive strength of 58.392 psi (pounds per square inch) with a standard deviation of s 648 psi. Test the null hypothesis H, -H = 58,000 psi against the alternative hypothesis H; H> 58,000 psi at 5% level of significance (value for t at 5 degree of freedom and 5% significance level is 2.0157). Here u denotes the population mean. (AMLE, Summer 2000)
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- TF.17 The average number of UHD TVs sold daily at a Best Buy store is known to be 28 and approximately normally distributed. A random sample of 21 days shows sample mean x̄ = 31 with standard deviation s = 7.7. Test the hypothesis Ho : μ = 28 against Ha : μ ≠ 28 at α = 0.10 and at α = 0.05 levels of significance. Use t-distribution.15Suppose in a local Kindergarten through 12th grade (K -12) school district, 49% of the population favor a charter school for grades K through 5. A simple random sample of 144 is surveyed. a. Find the mean and the standard deviation of X of B(144, 0.49). Round off to 4 decimal places. O = b. Now approximate X of B(144, 0.49) using the normal approximation with the random variable Y and the table. Round off to 4 decimal places. Y - N( c. Find the probability that at most 81 favor a charter school using the normal approximation and the table. (Round off to z-values up to 2 decimal places.) P(X 75) - P(Y > a (Z > e. Find the probability that exactly 81 favor a charter school using the normal approximation and the table. (Round off to z-values up to 2 decimal places.) P(X = 81) - P(Suppose in a local Kindergarten through 12th grade (K -12) school district, 49% of the population favor a charter school for grades K through 5. A simple random sample of 144 is surveyed. a. Find the mean and the standard deviation of X of B(144, 0.49). Round off to 4 decimal places. O = b. Now approximate X of B(144, 0.49) using the normal approximation with the random variable Y and the table. Round off to 4 decimal places. Y - N( c. Find the probability that at most 81 favor a charter school using the normal approximation and the table. (Round off to z-values up to 2 decimal places.) P(X 75) - P(Y > a (Z > e. Find the probability that exactly 81 favor a charter school using the normal approximation and the table. (Round off to z-values up to 2 decimal places.) P(X = 81) - P(J A survey found that women's heights are normally distributed with mean 63.3 in. and standard deviation 2.7 in. The survey also found that men's heights are normally distributed with mean 67.5 in. and standard deviation 3.4 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57 in. and a maximum o 64 in. Complete parts (a) and (b) below. ... a. Find the percentage of men meeting the height requirement. What does the result suggest about the genders of the people who are employed as characters at the amusement park? The percentage of men who meet the height requirement is 3.8%. (Round to two decimal places as needed.) an exampleA nutritionist claims that the mean tuna consumption by a person is 3.8 pounds per year. A sample of 70 people shows that the mean tuna consumption by a person is 3.5 pounds per year. Assume the population standard deviation is 1.04 pounds. At α=0.09, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. A. H0: μ=3.8 Ha: μ≠3.8 B. H0: μ≠3.5 Ha: μ=3.5 C. H0: μ>3.5 Ha: μ≤3.5 D. H0: μ>3.8 Ha: μ≤3.8 E. H0: μ≤3.5 Ha: μ>3.5 F. H0: μ≤3.8 Ha:μ>3.8 b) Identify the standardized test statistic. z= (Round to two decimal places as needed.) (c) Find the P-value. (d) Decide whether to reject or fail to reject the null hypothesis. A. Reject H0. There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.8pounds. B. Reject H0. There is sufficient evidence to reject the claim that mean tuna consumption is equal to…A nutritionist claims that the mean tuna consumption by a person is 3.9 pounds per year. A sample of 80 people shows that the mean tuna consumption by a person is 3.8 pounds per year. Assume the population standard deviation is 1.02 pounds. At a = 0.09, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. O A. Ho: u>3.8 Ha: μs 3.8 O B. Ho: H#3.8 Ha: μ= 3.8 O C. Ho: Hs3.9 Hai H> 3.9 O D. Ho: H>3.9 O E. Ho: Hs3.8 Ha: H> 3.8 F. Ho: H=3.9 Ha:us3.9 Ha: H#3.9 (b) Identify the standardized test statistic. (Round to two decimal places as needed.)A population has parameters u = 214.3 and o = 73.5. You intend to draw a random sample of size n = 237. What is the mean of the distribution of sample means? What is the standard deviation of the distribution of sample means? (Report answer accurate to 2 decimal places.) = 20A normal population sample has a standard deviation of 16 and sample size m = 100 . The rejection region for testing H0: ? = 5 ?????? Ha: ? = k is that X > k − 2. Find the value of k that will allow P(X > k− 2|H0: ? = 5 )=0.0228.Recommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th…StatisticsISBN:9781305251809Author:Jay L. 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