. A random sample of the ages of 41 young athletes has a mean of 20 years and a standard deviation of 2.5 years. Construct a 95% confidence interval estimate of the mean age of all young athletes.
Q: A random sample of 64 business students required an average of 100 minutes to complete a statistics…
A: Given that, Mean = 100 Standard deviation = 20 Sample size = 64 95% confidence level
Q: Insurance Company A claims that its customers pay less for car insurance, on average, than customers…
A: Claim: Insurance Company A claims that its customers pay less for car insurance, on average, than…
Q: what would the mean and standard deviation of the sample proportion ˆp be?
A: Let X be Pennsylvanians who have a high school diploma Given, n = 225 p = 0.83
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A: As the population standard deviation is known, we will use z distribution. The value of z at 95%…
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A: In question, Given that in a poll of 8 networking sites the average monthly visitor count was 12.1…
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Q: 2. A recent study of 25 students showed that they spent on average of $18.35 for gasoline per week.…
A: Given x̄ = 18.35 n=25 s=3 degree of freedom = df = n -1 = 25-1 = 24 significance level α = 0.10…
Q: The following sample data are from a normal population: 10, 8, 12, 15, 13, 11, 6, 5. What is the…
A: Given data is10,8,12,15,13,11,6,5sample size(n)=8sample mean(x¯)=10standard…
Q: A sample of 56 Charleston County households have a mean income of $27,519 with a standard deviation…
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Q: A car repair business randomly selects receipts for parts purchased on 16 different jobs. The mean…
A: given data, n=16x¯=31.32s=30.42CI=0.95α=1-0.95=0.05df=n-1=16-1=15we have to construct the 95%…
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…
Q: A recent study shows that out of 5, 000 people who moved to Arizona, 2, 350 did so to take a new…
A: From the given information, sample size n = 1000. Total number of people moved n(S) = 5000. Number…
Q: A sample of 47 Charleston County households have a mean income of $32,346 with a standard deviation…
A: The sample size is 47, sample mean is $32346, and sample standard deviation is $8571. The confidence…
Q: 2. A random sample of 85 group leaders, supervisors, and similar personal revealed that a person…
A: Introduction: The approach to determining the population distribution average is by utilizing a…
Q: A tester sampled randomly 150 bolts from a consignment. She noticed a standard deviation 6.2. Given…
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Q: Millionaires. Professor Thomas Stanley of Georgia State University has surveyed millionaires since…
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Q: In a survey, 45 people were asked how much they spent on their last partners birthday gift. The…
A: Given information- Sample size, n = 45 Significance level, α = 0.10 Sample mean, x-bar = $90 Sample…
Q: Insurance Company A claims that its customers pay less for car insurance, on average, than customers…
A: The question is about hypothesis testing.Given :Randomly selected no. of people who buy insurance…
Q: Calculate a 95% upper confidence bound for the mean of a normally distributed population using the…
A: Given that, x = 32289 s = 2846 n=15 For 95% confidence interval,the z value is 1.96
Q: Insurance Company A claims that its customers pay less for car insurance, on average, than customers…
A: Company A:Sample size Sample mean Sample standard deviation Company B:Sample size Sample mean Sample…
Q: The daily salaries of security guards for 10 security agencies are as follows: P350, P400, P375,…
A: Solution To find the point estimate we will find the sample mean.
Q: What is the source of variability in the random variable? O A. Random errors OB. The sample size C.…
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Q: the following random sample was selected from a normal distribution: 11, 16, 3, 13, 18, 18, 16, 19,…
A: Given data is11, 16, 3, 13, 18, 18, 16, 19, 17, 11sample size(n)=10
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: For this study, we should use t-test since the population standard deviation is unknown. Step…
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Q: A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each…
A: When the given sample size is less than 30 we use t-test as a test statistic. Using t-test we can…
Q: Considerr a samplee with data values of 10, 12, 16, 17 andd 20. The mean is 15 andd the sample…
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Q: n a survey, 25 people were asked how much they spent on their child's last birthday gift. The…
A: Givensample size(n)=25Mean(x)=38standard deviation(s)=19confidence interval(c)=98%
Q: The lead levels in a random sample of 35 crows in the region were measured and recorded. The mean…
A: It is given that Sample size n = 35 Sample mean M = 4.50 Sample SD s = 1.12 Confidence level = 99%…
Q: 2. One-Way Airfares The average one-way airfare from Pittsburgh to Washington, D.C., is $236. A…
A: Given:μ = 236n = 20x = 210s = 42a = 0.02 Defining the hypotheses:H0: μ = 236H1: μ ≠ 236 Test…
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A: The objective of this question is to determine the sample size needed to estimate the true mean…
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Q: We take a simple random sample of 16 adults and ask them how long they sleep on atypical night. The…
A: The objective of this question is to calculate the 88% confidence interval for the true mean time…
Q: 73 full time workers were surveyed on how many hours they spend watching television during the week.…
A: Given information, n=73x¯=3.2s=0.2 Confidence level is 95% significance level is 5% df=n-1=73-1=72 t…
Q: Which type of confidence interval should you use in this situation? An economics professor randomly…
A: n=sample size=100, X¯=54.8,s=7.9
Q: A random sample of 30 words from Jane Austen's Pride and Prejudice had a mean length of 4.08 letters…
A: Given Jane Austen's P&P Sample size of words, n1=30 Mean length, x1=4.08 Standard Deviation,…
Q: An article in the San Jose Mercury News stated that students in the California state university…
A: The population mean is μ=4.5 The sample mean and sample standard deviation is x=5.1sd=1.2n=49
Q: A high school athletic director is asked if basketball players are doing as well academically as the…
A: Given that Sample size n =20 Sample mean =3.18 Sample standard deviation =0.54
Q: company manufactures a certain over-the-counter drug. The company samples 80 pills and finds that…
A: We have given that Sample size n = 80 Sample mean = 325.5 Standard deviation s=10.3 90%confidence…
Q: Insurance Company A claims that its customers pay less for car insurance, on average, than customers…
A: The insurance Company A claims that its customers pay less for car insurance, on average, than…
Q: Insurance Company A claims that its customers pay less for car insurance, on average than customers…
A: Formula : test statistic for equal variance t test is
Q: You wish to evaluate the differences in educational quality between two training centres (A and B).…
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- 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 13 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $19. For 9 randomly selected customers of Company B, you find that they pay a mean of $157 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.05 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.At the local fish hatchery we take a random sample of 21 trout from one of the holding pens. The trout in the sample have a mean length of 11.2 inches with a standard deviation of .73 inches. Construct and interpret a 95% confidence interval for the mean length of all trout in this holding pen. Type your interpretation sentence in the written response box.In a simple random sample of 144 households in a county in Virginia, the average number of children in these households was 3.62 children. The standard deviation from this sample was 2.40 children. A 90% confidence interval for the mean number of children in all households in this county is?
- According to the College Board, scores on the math section of the SAT Reasoning college entrance test for the class of 2010 had a mean of 516 and a standard deviation of 116. Assume that they are roughly normal.One of the quartiles of the scores from the math section of the SAT Reasoning test is 438. The other quartile is _______.1. For a group of 10 students subjected to a stress situation the mean number of heartbeats/minutes was 126 and the standard deviation was 4. Find the 95% confidence interval for the true mean.2. A sample of six college basketball player had an average weight of 217 pounds with a sample standard deviation of 10 pounds. Find the 95% confidence interval of the true mean weight of all basketball player.
- 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 13 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $19. For 9 randomly selected customers of Company B, you find that they pay a mean of $157 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho: M₁ = μ₂ Ha:M₁ •H₂Insurance 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 15 people who buy insurance from Company A, the mean cost is $154 per month with a standard deviation of $13. For 11 randomly selected customers of Company B, you find that they pay a mean of $159 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.02 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.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 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 _____…Toby wants to know the average price of new cars at a particular dealearship. He randomly selects a sample of 31 cars and finds that the average price to be $19500$19500 with a standard deviation of $1640$1640. What is the 95% confidence interval for the mean price of cars at this particular dealership? With 95% confidence, we can say the true population mean price of cars at this dealership is between $____ and _____$ .Round to the nearest dollar