Table below shows the observed frequency of the number of received messages by an air traffic control of Subang airport. TEST using a = 0.05, whether the sample is from a population that has a Poisson distribution with µ = 4.6 (CO3, C4, PO2) No. of radio Observed messages frequency 3 1 15 2 47 76 4 68
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- of 21 Page 4. Bluethroat Song Duration Male bluethroats have a complex song which is thought to be used to attract female birds. Let x denote the duration of a randomly selected song (in seconds) from a male bluethroat. The authors of research on bluethroat song report the mean song duration is 13.8 seconds and the standard deviation of song durations is 11.8 seconds. The authors also noted that the song length distribution is not normal. a. Let X = average song duration (in seconds) for a sample of 36 male bluethroat songs. Is this distribution of the sample mean song duration "normally distributed"? Explain how you know. X = average song duration (in seconds) for the sample of 36 male bluethroat songs. X- is distributed according to a distribution with u = b. Find the probability that a sample of 36 male bluethroat songs will have a mean duration greater than12 seconds. Draw, label, and shade a graph to illustrate your result. Find P( ) DThe breaking strengths of cables produced by a certain manufacturer have historically had a mean of 1875 pounds and a standard deviation of 70 pounds. The company believes that, due to an improvement in the manufacturing process, the mean breaking strength, µ, of the cables is now greater than 1875 pounds. To see if this is the case, 46 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1886 pounds. Assume that the population is normally distributed. Can we support, at the 0.05 level of significance, the claim that the population mean breaking strength of the newly- manufactured cables is greater than 1875 pounds? Assume that the population standard deviation has not changed. Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.) (a) State the null hypothesis H and the…The breaking strengths of cables produced by a certain manufacturer have historically had a mean of 1775 pounds and a standard deviation of 70 pounds. The company believes that, due to an improvement in the manufacturing process, the mean breaking strength, u, of the cables is now greater than 1775 pounds. To see if this is the case, 14 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1812 pounds. Assume that the population is normally distributed. Can we support, at the 0.10 level of significance, the claim that the population mean breaking strength of the newly- manufactured cables is greater than 1775 pounds? Assume that the population standard deviation has not changed. Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.) (a) State the null hypothesis H, and the…
- 8. 140 migrating pigeons were caught by a biologist for data collection. The mass of these pigeons is normally distributed with mean 0.9 kg and standard deviation of deviation 0.15 kg. a) Determine the percentile rank of a pigeon weighing 1kg. b) What proportions of pigeons have weight greater than 1.1 kg or less than 0.7 Kg 31In a trivariate distribution, it 112= 0.7, 113 = 0.61 and 223 = 0.4 find all the multiple correlation coefficients. Also obtain the standard of estimates 01.23 61:23, 02:13 and 53:12 given that: 52= 1:06, 62 = 2.15 and 3= 3.46.The average American gets a haircut every 43 days. Is the average larger for college students? The data below shows the results of a survey of 11 college students asking them how many days elapse between haircuts. Assume that the distribution of the population is normal. 38, 37, 38, 45, 53, 55, 35, 37, 35, 41, 38 What can be concluded at the the α = 0.05 level of significance level of significance? For this study, we should use? Select an answer: t-test for a population mean? z-test for a population mean? The null and alternative hypotheses would be: H0:H0? p or μ Select an answer > = ≠ ≤ < ≥ H1:H1? μ or p Select an answer ≠ ≥ > ≤ = < 2. The test statistic ? t z = (please show your answer to 3 decimal places.) The p-value = _______ (Please show your answer to 3 decimal places.) The p-value is? > or ≤ α 3. Based on this, we should?? Select an answer reject ? accept ? or fail to reject ?…
- Steel rods manufactured in a plant must be a minimum of 2 metres in length in order to pass through an inspection procedure. Steel rods with lengths that are normally distributed, with a mean of 2.1 metres and a standard deviation of 0.1 metres, are produced during the manufacturing process. A random sample of 25 steel rods is chosen, and their lengths are measured using a tape measure. i. Decide on the appropriate distribution of the sample mean of the 25 steel rods in the sample using the appropriate distribution formula. ii. Calculate the probability that the sample mean of steel rods that meet the minimum length requirement is the same as the mean of the entire sample. iii. For example, if the observed sample mean length of steel rods is 2.15, calculate the 98% two-sided confidence interval for the true mean length of the manufacturing process, assuming that it is not known.let x be a normal ramdom variable with a mean =169.5 and standard deviation =4.2 ramdom samples of fixed size n=30 are drawn from the distribution, mean, standard deviation od sample meanDo 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…
- The breaking strengths of cables produced by a certain manufacturer have historically had a mean of 1775 pounds and a standard deviation of 70 pounds. The Español company believes that, due to an improvement in the manufacturing process, the mean breaking strength, µ, of the cables is now greater than 1775 pounds. To see if this is the case, 16 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1806 pounds. Assume that the population is normally distributed. Can we support, at the 0.10 level of significance, the claim that the population mean breaking strength of the newly- manufactured cables is greater than 1775 pounds? Assume that the population standard deviation has not changed. Perform a one-tailed test. Then complete the parts below. 00 Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.) (a) State the null hypothesis H,…What is the sample correlation coefficient for these data? Carry your intermediate computations to at least four decimal places and round your answer to at least three decimal places.The average house has 13 paintings on its walls. Is the mean different for houses owned by teachers? The data show the results of a survey of 13 teachers who were asked how many paintings they have in their houses. Assume that the distribution of the population is normal. 10, 13, 13, 14, 12, 12, 13, 11, 14, 11, 12, 14, 11 What can be concluded at the αα = 0.05 level of significance? For this study, we should use Select an answer z-test for a population proportion t-test for a population mean The null and alternative hypotheses would be: H0:H0: ? μ p Select an answer < ≠ > = H1:H1: ? μ p Select an answer = > ≠ < The test statistic ? z t = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? ≤ > αα Based on this, we should Select an answer fail to reject accept reject the null hypothesis. Thus, the final conclusion is that ... The data suggest the populaton mean is…