A tire manufacturer has been producing tires with an average life expectancy of 28,000 miles. Now the company is advertising that its new tires' life expectancy has increased. In order to test the legitimacy of the advertising campaign, an independent testing agency tested a sample of 12 of their tires and has provided the following data.
Q: A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A: Given,
Q: A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A: Given Population mean μ=707.7, n=sample size=31, sample mean x̄=725.2, sample standard deviations…
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A: Note: Thank you for the question. Since multiple subparts are posted, according to our policy, we…
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A: It is needed to determine the dependent and independent variable.
Q: A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A: The test statistic is 1.59 and the P-value is 0.060.
Q: A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A:
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A: Two sample proportion test: Two sample proportion test is used to test the significance difference…
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A: The objective is to find the required probabilities.
Q: A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A: Denote μ as the true mean credit score.
Q: Sleep apnea is a condition in which the sufferers stop breathing momentarily while they are asleep.…
A: Let p be the population proportion p=5.9%=0.059 Let X be the number of people who suffer from…
Q: State the null and alternative hypothesis as an expression
A: Given that x1=12.5 x2=13.9 s1=3.9 s2=5.2 n1=14 n2=17
Q: score A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A: Given:Mean(x)=716.2Standard deviation(s)=81.1Sample size(n)=34α=0.05
Q: Sleep apnea is a condition in which the sufferers stop breathing momentarily while they are asleep.…
A: Claim : the percentage of commercial truck drivers who suffer from sleep apnea is not 4.1%
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A: Given that, Sample size n=22 It was desired to test to determine if that average price of all…
Q: A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A: Population mean, Sample size, Sample mean, Sample standard deviation, s = 81.1
Q: A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A: According to a survey, the mean credit score is 700.8 A random sample of 45 high-income…
Q: A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A:
Q: A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A: The sample size is 45, the sample mena is 725.6 and the sample standard deviaiton is 81.6.
Q: Morgan City Fire Chief Sidney Pyro is concerned about the low efficiency scores that his…
A: A two-sample independent t-test is performed to test if there is a statistically significant…
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A: Given: n= 2300 ,x= 805 p= 0.33 ,significance level(∝)= 95.0 p̂= 805 / 2300 =define hypothesis;-
Q: Sleep apnea is a condition in which the sufferers stop breathing momentarily while they are asleep.…
A: We have to find test statastics.
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A: Solution: From the given information, the mean of the differences is 1.33 and the standard deviation…
Q: Sleep apnea is a condition in which the sufferers stop breathing momentarily while they are asleep.…
A: Givenx=19n=308p^=19308=0.0617α=0.05
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A: 1.The hypotheses for the provided test can be written as:
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Q: A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A: One sample t-test: One sample t-test is used to test the significance difference between population…
Q: A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A: One sample t test: One sample t test is the hypothesis test that is used to test whether the mean of…
Q: A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A: Population mean, Sample size, Sample mean, Sample standard deviation, s = 81.1
Q: A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A: Given that ; u = 700.4 Sample mean = 716.2 n = 34 s = 81.1 Here we use one sample mean test to…
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A: In hypothesis testing, a pre-defined value of the population parameter is tested. To do this, a…
Q: A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A: Let us consider a population follows a normal distribution Let a sample be taken from that…
Q: A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A: There is always some contention about the values of a parameters. When parametric values are unknown…
Q: A consulting firm had predicted that 30% of employees at a large firm would take advantage of a new…
A: From the provided information, A consulting firm had predicted that 30% of employees at a large firm…
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A: Testing of Hypothesis: Testing of hypothesis is a rule which, when the sample values have been…
Q: Tourism is extremely important to the economy of Florida. Hotel occupancy is an often-reported…
A: a. Assume that p1, p2 are the true population proportions of occupied rooms in current year and…
Q: A company claims that their product has an average lifespan of 6 years. An independent company…
A: It is an important part of statistics. It is widely used.
Q: A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A: Null hypothesis is the hypothesis of no difference. Alternative hypothesis is the claim being made…
Q: A credit score is used by credit agencies (such as mortgage companies and banks) to assess the…
A: Given data : sample size, n =34 sample mean, x̄ = 716.2 sample standard…
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- A credit score is used by credit agencies (such as mortgage companies and banks) to assess the creditworthiness of individuals. Values range from 300 to 850, with a credit score over 700 considered to be a quality credit risk. According to a survey, the mean credit score is 703.8. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher credit scores. He obtained a random sample of 36 high-income individuals and found the sample mean credit score to be 719.8 with a standard deviation of 81.9. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the α=0.05 level of significance.According to the American Lung association, lung cancer account for 28% of all cancer deaths in the US. A new type of screening for lung cancer, computed tomography (CT), has been developed. Medical researchers believe CT scans are scans are more sensitive than regular X-rays in pinpointing small tumors. The Cancer Center at the University of South Florida is currently conducting a clinical trial of 50,000 smokers nationwide to compare the effectiveness of CT scans with X-ray for detecting lung cancer (Today’s Tomorrows, Fall 2002). Each participating smoker is randomly assigned to one of the two screening methods, CT or chest X-ray, and his or her progress is tracked over time. The age at which the scanning method first detects a tumor is the variable of interest. A. Identify the data collection method used by the cancer researchers. B. dentify experimental units of the study. C. Identify the type of the variable measured. D. Identify the population and sample.A credit score is used by credit agencies (such as mortgage companies and banks) to assess the creditworthiness of individuals. Values range from 300 to 850, with a credit score over 700 considered to be a quality credit risk. According to a survey, the mean credit score is 708.2. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher credit scores. He obtained a random sample of 31 high-income individuals and found the sample mean credit score to be 723.3 with a standard deviation of 80.9. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the α = 0.05 level of significance. C State the null and alternative hypotheses. Ho: HY H₁: (Type integers or decimals. Do not round.) Identify the t-statistic. to = (Round to two decimal places as needed.) Identify the P-value. P-value = (Round to three decimal places as needed.) Make a conclusion regarding the hypothesis. the null hypothesis.…
- A credit score is used by credit agencies (such as mortgage companies and banks) to assess the creditworthiness of individuals. Values range from 300 to 850, with a credit score over 700 considered to be a quality credit risk. According to a survey, the mean credit score is 700.1. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher credit scores. He obtained a random sample of 45 high-income individuals and found the sample mean credit score to be 713.4 with a standard deviation of 84.9. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the a = 0.05 level of significance. State the null and alternative hypotheses. Ho: H (Type integers or decimals. Do not round.) Identify the t-statistic. to = (Round to two decimal places as needed.) Identify the P-value. P-value = (Round to three decimal places as needed.) e%3D Make a conclusion regarding the hypothesis. the null hypothesis. There…A credit score is used by credit agencies (such as mortgage companies and banks) to assess the creditworthiness of individuals. Values range from 300 to 850, with a credit score over 700 considered to be a quality credit risk. According to a survey, the mean credit score is 704.5. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher credit scores. He obtained a random sample of 36 high-income individuals and found the sample mean credit score to be 714.9 with a standard deviation of 81.7. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the a = 0.05 level of significance. State the null and alternative hypotheses. Ho: H V H1: H V (Type integers or decimals. Do not round.) Identify the t-statistic. to (Round to two decimal places as needed.) Identify the P-value. P-value = (Round to three decimal places as needed.) Make a conclusion regarding the hypothesis. V the null hypothesis.…Genetic drugs are lower cost substitutes for brand-name drugs. Before a genetic drug can be sold in the United States, it must be tested found to perform equivalently to the brand-name product. The U.S. Food and Drug Administration is now supervising the testing of new genetic antifungal ointment. The Brand-name ointment is known to deliver a mean of 3.5 micrograms of active ingredient to each square centimeter of skin. As part of the testing, seven subjects apply the ointment. Six hours later, the amount of drugs that has been absorbed into the skin measured. The amounts, in micrograms, are 2.6, 3.2, 2.1, 3.0, 3.1, 2.9, 3.7 How strong is the evidence that the mean amount absorbed differs from 3.5 micrograms? Test by using a confidence interval with ?=0.05α=0.05. Show Step1-Step3. Round critical value to nearest hundredth. (e.g. 0.1256 would be entered as 0.13) Round confidence interval to nearest thousandth. (e.g. 0.12456 would be entered as 0.125)
- A credit score is used by credit agencies (such as mortgage companies and banks) to assess the creditworthiness of individuals. Values range from 300 to 850, with a credit score over 700 considered to be a quality credit risk. According to a survey, the mean credit score is 702.4. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher credit scores. He obtained a random sample of 41 high-income individuals and found the sample mean credit score to be 721.3 with a standard deviation of 80.9. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the a= 0.05 level of significance. State the null and alternative hypotheses. Ho H H₁ H (Type integers or decimals. Do not round.) Identify the t-statistic. to = (Round to two decimal places as needed.) Identify the P-value. P-value= (Round to three decimal places as needed.) Make a conclusion regarding the hypothesis. the null hypothesis. There…A credit score is used by credit agencies (such as mortgage companies and banks) to assess the creditworthiness of individuals. Values range from 300 to 850, with a credit score over 700 considered to be a quality credit risk. According to a survey, the mean credit score is 704.9. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher credit scores. He obtained a random sample of 33 high-income individuals and found the sample mean credit score to be 717.9 with a standard deviation of 83.9. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the g = 0.05 level of significance. State the null and alternative hypotheses. Ho: µ 704.9 H1: µ > 704.9 (Type integers or decimals. Do not round.) Identify the t-statistic, to = (Round to two decimal places as needed.)A credit score is used by credit agencies (such as mortgage companies and banks) to assess the creditworthiness of individuals. Values range from 300 to 850, with a credit score over 700 considered to be a quality credit risk. According to a survey, the mean credit score is 705.8. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher credit scores. He obtained a random sample of 37 high-income individuals and found the sample mean credit score to be 721.3 with a standard deviation of 81.9. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the a = 0.05 level of significance. State the null and alternative hypotheses. Họ: H H1: H (Type integers or decimals. Do not round.)
- What percentage of Contra Costa County adults supports a ban on assault-style weapons? Suppose that we survey a random sample of 100 adults living in Contra Costa County and find that 62% support a ban on assault-style weapons. According to a 2015 study by the Pew Research Center, 57% of U.S. adults favor a ban on assault-style weapons. Assuming that the variability in random samples will be the same in Contra Costa County as in the U.S., find the 95% confidence interval to estimate the proportion of CCC adults that favor a ban on assault-style weapons. g. Are we confident that the percentage of Contra Costa County residents that supports a ban is greater than the percentage nationwide as reported by the Pew Research Center? Why or why not?Dr. Blake is interested in the effects of intoxication on driving ability. Participants are recruited and encouraged to consume one to four mixed drinks during a one-hour period. Dr. Blake records the number of driving errors made while the participants navigate a 20-mile course in a driving simulator and measures the participants' blood alcohol content at the end of the simulation. What statistical test should Dr. Blake use to analyze these data? a. one-way within-groups ANOVA b. one-way between-groups ANOVA c. chi-square test of indepandence d. Pearson correlation coefficientSleep apnea is a condition in which the sufferers stop breathing momentarily while they are asleep. This condition results in lack of sleep and extreme fatigue during waking hours. A current estimate is that 16.3 million out of the 312.7 million Americans suffer from sleep apnea, or approximately 5.2%. A safety commission is concerned about the percentage of commercial truck drivers who suffer from sleep apnea. They do not have any reason to believe that it would be higher or lower than the population’s percentage. To test the claim that the percentage of commercial truck drivers who suffer from sleep apnea is not 5.2%, a simple random sample of 380380 commercial truck drivers is examined by a medical expert, who concludes that 29 suffer from sleep apnea. Does this evidence support the claim that the percentage of commercial truck drivers who suffer from sleep apnea is not 5.2%? Use a 0.01 level of significance. Step 1 of 3 : State the null and alternative hypotheses for the test.…
