Example 9: Find the 95% confidence interval for the variance and standard deviation of the nicotine content of cigarettes manufactured if a sample of 20 cigarettes has a standard deviation of 1.6 milligrams.
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A: Solution: State the hypotheses. Null hypothesis: That is, there is no difference in the variation…
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A: Given that, Mean = 100 Standard deviation = 20 Sample size = 64 95% confidence level
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Q: Insurance Company A claims that its customers pay less for car insurance, on average, than customers…
A: Answer Given Mean [x1] =151 Mean [x2] =158 Standard deviation [s1] =19 Standard deviation [s2] =14…
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A: From the provided information, Sample size (n) = 12 Standard error (SE) = 0.4
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A: a) Confidence interval: The (1-α) % confidence interval for population variance is given by:
Q: The standard deviation of a distribution is hypothesized to be 50. If an observed sample of 16…
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Q: Show that X is a sufficient estimator of the mean u of a normal population with the known variance…
A: X~N(μ,σ2)
Q: Find the 97% confidence interval for the variance and standard deviation for the time it takes a…
A: In question, We'll find the 97% confidence interval for the variance and standard deviation for the…
Q: Death row and the single-sample t test: The Florida Department of Corrections publishes an online…
A: Solution a. The following data is provided in the question 25.62, 13.09, 8.74, 17.63, 2.8, 4.42,…
Q: A. Mean = 1.445 %3D B. Variance = 22.603 C. Standard Deviation = %3D
A: X p(x) xp(x) (x-xbar)^2 p(x)*(x-xbar)^2 -4 0.573 -2.292 29.64803 16.98832 7 0.149 1.043 30.85803…
Q: 10. Is it possible for the standard deviation of a list of numbers to equal zero? If so, give an…
A: Q10. Numbers can be repeated
Q: The Water Resource Department claims that the mean pH level of the river water is 6.8. You randomly…
A: Given,sample size(n)=19sample mean(x¯)=6.7sample standard deviation(s)=0.24population…
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A: Given that Sample sizes n1=70 , n2=85 Sample means X̄1=8.84 , X̄2=9.05 ?1 =0.30…
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A: The provided information is For Ax¯1=11s1=3.6x¯2=15s2=3.6n1=15n2=23α=0.10 a. The null and…
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Q: Draw a conclusion and interpret the decision
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Q: ize 400 is taken from a population with a mean of 610 and a standard deviation of 15. Find the…
A: Given, n=400 Mean (μ) = 610 Standard deviation (σ) = 15
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Q: Question: A sample of 25 items has a mean of 45 and a standard deviation of 5. What is the standard…
A: Introduction: The standard error of the mean (SEM) is a measure of the variability of the sample…
Q: A medical researcher administers a new medication to a random sample of 70 flu sufferers, and she…
A: From the provided information, The degree of freedom = n1 + n2 – 2 = 70 + 50 – 2 = 118 As, the…
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A: There are two samples of species which are named as species A and species B. The species follows…
Q: A sample has n = 16 scores, a mean of M = 45 and has an estimated standard error of 4 points (sM =…
A: Given information is A sample has n = 16 scores, a mean of M = 45 and has an estimated standard…
Q: Company ABC's customer retentions per site are normally distributed with a standard deviation of 21…
A: Solution: Let X be the Company ABC’s customer retentions per site. From the given information, X…
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A: a) Consider that the population standard deviation of the ages of the Playstation users σ1 and the…
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A: Standard Deviation: Standard deviation measures how dispersed your data in from the mean.
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A: The objective is to find range, variance and standard deviation for the given sample data.
Q: Find the critical value(s). Round the answer to at least two decimal places. If there is more than…
A: The level of significance is 0.10. Computation of critical value: α=0.10α2=0.051-α2=0.95 The…
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Q: Consider two populations in the same state. Both populations are the same size (22,000). Population…
A: The following solution is provided below :
Q: You would like to estimate the variance of GPA’s at SDSU. You randomly select 14 students and get a…
A: From the provided information, Sample size (n) = 14 Sample standard deviation (s) = 0.981 Confidence…
Q: Moviegoers The average "moviegoer" sees 8.5 movies a year. A moviegoer is defined as a person who…
A: Given Xbar=9.1 , sigma=3.2 , n=47 Alpha=0.10 H0:mu=8.5 Vs H1:mu≠8.5 Critical value=-1.645 to…
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A: Given: n = 29 x¯ = 597.9 s = 3.8 α = 0.01
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Q: Scores on an IQ test are normally distributed. A sample of 18 IQ scores had standard deviation s =…
A: Confidence interval: The (1-α) % confidence interval for population variance is given by:
Q: Calculate the observed z value using a sample mean of 25, a population average of 20, and a standard…
A: Given data: Sample mean = 25 Population average = 20 Standard error of the mean = 8
Q: Gestation Periods of Humans. For humans, gestation periods are normally distributed with a mean of…
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- One-Way Airfares The average one-way airfare from Pittsburgh to Washington D.C. is $236. A random sample of 19 one-way fares during a particular month had a mean of $208 with a standard deviation of $43. At α=0.05, is there sufficient evidence to conclude a difference from the stated mean?Two accounting professors decided to test if the variance of their grades is different or no. To accomplish this, they each graded the same 10 exams with the following results: Standard Deviation 22.4 Mean Grade Professor 1 79.3 Professor 2 82.1 12.0 F What is the P-value of the test? Select one: a. between 0.01 and 0.025 Ob. between 0.025 and 0.05 C. between 0.05 and 0.10 Od. bigger than 0.10Fear of heights - A psychologist has developed a new treatment for acrophobia (the extreme fear of heights) and wants to compare the results of this new treatment to the results of the standard treatment. A study is conducted where a random sample of 16 patients undergo the new treatment and their improvement scores on a diagnostic test are compared to a control sample of 13 patients who received the standard treatment. Summary scores for both groups are shown in the table below. Treatment Sample Mean Sample Standard Deviation New 50.9 12 Standard 47.2 11.4 The degrees of freedom for this problem is d f 26.287709. 1. Select the hypotheses that should be used to assess if the new treatment results in higher average improvement scores than the standard treatment. А. Но : И1 = µ2 vs. Ha : H1 # µ2 OB. Ho : µ1 = µ2 vS. Ha : H1 H2 2. Calculate the test statistic. ? V = 3. Calculate the p-value. p-value =
- Choose the appropriate statistical test. When computing, be sure to round each answer as indicated. A dentist wonders if depression affects ratings of tooth pain. In the general population, using a scale of 1-10 with higher values indicating more pain, the average pain rating for patients with toothaches is 6.8. A sample of 30 patients that show high levels of depression have an average pain rating of 7.1 (variance 0.8). What should the dentist determine? 1. Calculate the estimated standard error. (round to 3 decimals). [st.error] 2. What is thet-obtained? (round to 3 decimals). 3. What is the t-cv? (exact value) 4. What is your conclusion? Only type "Reject" or Retain"A report stated that after obtaining a bachelor's degree, 43% of college students enroll in a graduate degree program. For a random sample of 61 undergraduate students, find the mean, variance and standard deviation for the number of students who will enroll in a graduate degree program after obtaining a bachelor's degree. a.) mean b.) variance c.) standard deviation (round to two decimal places)Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 1313 people who buy insurance from Company A, the mean cost is $151$151 per month with a standard deviation of $16$16. For 99 randomly selected customers of Company B, you find that they pay a mean of $158$158 per month with a standard deviation of $19$19. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.050.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.
- A manufacturer advertises that the average life of batteries produced by his firm is at least 12 months. The consumer board disagrees, and claims that the average life of the batteries is less than 12 months. The consumer board test a random sample of 9 batteries and find that the mean life is 10.6 months with a standard deviation of 2. The consumer board's alternative hypothesis is: A. The average life of the population of batteries is 12 months. B. The average life of the population of batteries is less than 12 months. C. The average life of the population of batteries is more than 12 months. D. The average life of the population of batteries is at least 12 monthsThe manufacturer of 'Road King' tyres claim that their tyres last an average of 70,000km or more. The Consumer Board suspects that this claim may be false (they have received some complaints) and decide to test the manufacturer's claim. The Consumer Board took a sample of size 500 from the population. They ran the…A Two-Sample Hypothesis TestThe mean number of English courses taken in a two-year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an average of three English courses with a standard deviation of 0.8. The females took an average of four English courses with a standard deviation of 1.0. Are the means statistically the same? (Assume a 5% level of significance.)Useful tools:Normal Distribution Calculatort-Distribution Calculator 1. Which of the following null and alternative hypotheses match this scenario? A---H0:Δμ=0H0:Δμ=0Ha:Δμ≠0Ha:Δμ≠0 B---H0:Δμ≠0H0:Δμ≠0Ha:Δμ=0Ha:Δμ=0 C---H0:Δμ=0H0:Δμ=0Ha:Δμ>0 2. Which type of test should be applied? A---The alternative hypothesis indicates a right-tailed test. B---The alternative hypothesis indicates a left-tailed test. C---The alternative hypothesis indicates a two-tailed test. 3. Which type of distribution should be…In a study of birth order and intelligence, IQ tests were given to 18- and 19-year-old men to estimate the size of the difference, if any, between the mean IQs of firstborn sons and secondborn sons. The following data for 10 firstborn sons and 10 secondborn sons are consistent with the means and standard deviations reported in the article. It is reasonable to assume that the samples come from populations that are approximately normal. Can you conclude that the mean IQ of firstborn sons is greater than the mean IQ of secondborn sons? Let μ1 denote the mean IQ of firstborn sons and μ2 denote the mean IQ of secondborn sons. Use the α = 0.01 level and the P-value method with the table. Firstborn 128 101 128 112 121 105 122 98 106 108 Secondborn 121 125 110 107 114 93 80 94 91 83 Part(a): State the appropriate null and alternate hypotheses. H0: H1: This is a _____…
- An intercity bus company is attempting to improve the reliability of its service. One manager claims that the variability (i.e., standard deviation) in arrivals from the posted schedule at the central bus depot is approximately 3 minutes. Another manager disagrees, suspecting that the spread in times of arrival is different. Using variance, what is the population parameter under consideration? What hypotheses should the managers test in this situation?13.34 Women's heights. The heights of adult women are approximately Normal with mean 69.2 inches and standard deviation 2.5 inches. How tall are the tallest 10% of women? (Use the closest percentile that appears in Table B.)
