At the 0.10 level of significance, can it be concluded that this represents a difference from the national average?
Q: Fran is training for her first marathon, and she wants to know if there is a significant difference…
A: The objective of this question is to test the claim that there is a significant difference between…
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Q: A parenting magazine reports that the average amount of wireless data used by teenagers each month…
A: The question is about hypo. testing Given : Popl. mean amount of wireless data used by teenagers ( μ…
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Q: Fran is training for her first marathon, and she wants to know if there is a significant difference…
A: Let be the group runners training for marathons be population 1 and be the individual runners…
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Q: Fran is training for her first marathon, and she wants to know if there is a significant difference…
A: Sample mean (x̅1) = 42.8Sample mean (x̅2) = 40.9Sample size (n1) = 34Sample size (n2) = 41Standard…
Q: Fran is training for her first marathon, and she wants to know if there is a significant difference…
A: The objective of this question is to test the claim that there is a significant difference between…
Q: A parenting magazine reports that the average amount of wireless data used by teenagers each month…
A: The objective of this question is to test Ella's claim that the average amount of wireless data used…
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A: Given S1=standard deviation form first sample=1.6S2=standard deviation form 2nd sample=2.1n1=first…
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A: Given information- Population mean, μ = $53 Sample size, n = 100 Sample mean, x-bar = $60 Sample…
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A: From the provided information, Mean (µ) = 17.2 Standard deviation (σ) = 3.4 X~N (17.2, 3.4)
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A: Givensample size(n)=49Mean(x)=5.1standard deviation(s)=1.2significance level(α)=0.01
Q: A marketing researcher wants to estimate the mean amount spent per year ($) on a web site by…
A: The summary of statistics is,
Q: Fran is training for her first marathon, and she wants to know if there is a significant difference…
A: Solution
Q: A parenting magazine reports that the average amount of wireless data used by teenagers each month…
A: The objective of the question is to state the null and alternative hypotheses for the test. The null…
Q: A parenting magazine reports that the average amount of wireless data used by teenagers each month…
A: The objective of this question is to test Ella's claim that the average amount of wireless data used…
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A: From the given information,
Q: A parenting magazine reports that the average amount of wireless data used by teenagers each month…
A: given data, population distribution is approximetly normal.α=0.005n=16claim:μ<5x¯=4.6s=1.5test…
Q: A parenting magazine reports that the average amount of wireless data used by teenagers each month…
A: From the provided information, Sample size (n) = 4 Sample mean (x̄) = 9.2 Sample standard deviation…
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Q: A parenting magazine reports that the average amount of wireless data used by teenagers each month…
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A: Given information Hypothesized mean µ = $49 Sample size (n) = 130 Mean x̅ = $55 Standard deviation…
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A: Let, X = Amount spent per year ($) on a web site by membership member shoppers. Given that;…
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A: Method By using the 90% confidence level to construct a confidence interval.
Q: A parenting magazine reports that the average amount of wireless data used by teenagers each month…
A: The objective of this question is to compute the value of the test statistic for Ella's claim that…
Q: Fran is training for her first marathon, and she wants to know if there is a significant difference…
A: From the provided information,
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A: given data n = 70 x¯ = 28 poundsσ = 3.8 poundsbest point estimate for population mean = ?
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A: Given data,n1=40x1=46.6σ1=4.6n2=45x2=48.8σ2=2.4Compute value of test statistic?
Q: parenting magazine reports that the average amount of wireless data used by teenagers each month is…
A: To test Ella's claim, we can perform a one-sample t-test. Here are the steps involved: Step 1:…
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Q: Fran is training for her first marathon, and she wants to know if there is a significant difference…
A: Let be the group runners training for marathons be population 1 and be the individual…
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Q: A marketing researcher wants to estimate the mean amount spent per year ($) on a web site by…
A: Given values, n=100x¯=60s=55
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A: 0.33271Explanation:For the given normal distribution with μ=26.8 and σ=7.4, the required proportion…
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A: The detailed solution is as follows below:
The average "moviegoer" sees 8.5 movies a year. A moviegoer is defined as a person who sees at least one movie in a theater in a 12-month period. A random sample of 50 moviegoers from a large university revealed that the average number of movies seen per person was 9.2. The population standard deviation is 3.2 movies. At the 0.10 level of significance, can it be concluded that this represents a difference from the national average?
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- Fran is training for her first marathon, and she wants to know if there is a significant difference between the mean number of miles run each week by group runners and individual runners who are training for marathons. She interviews 37 randomly selected people who train in groups, and finds that they run a mean of 47.7 miles per week. Assume that the population standard deviation for group runners is known to be 3.3 miles per week. She also interviews a random sample of 49 people who train on their own and finds that they run a mean of 49.4 miles per week. Assume that the population standard deviation for people who run by themselves is 4.4 miles per week. Test the claim at the 0.10 level of significance. Let group runners training for marathons be Population 1 and let individual runners training for marathons be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.Average adult Americans are about one inch taller but nearly a whopping 25 pounds heavier than they were in 1960, according to a report from the Centers for Disease Control and Prevention (CDC). City A is considered one of America's healthiest cities. Is the weight gain since 1960 similar in city A? A sample of n=25 adults suggested a mean increase of 22 pounds with a standard deviation of 9.4 pounds. Is City A significantly different in terms of weight gain since 1960? Complete and interpret 5 step hypothesis testing by hand at a 5% level of significance using a Z-test. Show each step and your work.A marketing researcher wants to estimate the mean amount spent per year ($) on a web site by membership member shoppers. Suppose a random sample of 100 membership member shoppers who recently made a purchase on the web site yielded a mean amount spent of $55 and a standard deviation of $52. Complete parts (a) and (b) below. a. Is there evidence that the population mean amount spent per year on the web site by membership member shoppers is different from $51? (Use a 0.01 level of significance.) State the null and alternative hypotheses. Ho: H (Type integers or decimals. Do not round. Do not include the $ symbol in your answer.) Identify the critical value(s). The critical value(s) is/are (Type an integer or a decimal. Round to two decimal places as needed. Use a comma to separate answers as needed.) Determine the test statistic. The test statistic, 1STAT, is (Type an integer or a decimal. Round to two decimal places as needed.) State the conclusion. Ho. There is evidence that the…
- Fran is training for her first marathon, and she wants to know if there is a significant difference between the mean number of miles run each week by group runners and individual runners who are training for marathons. She interviews 42 randomly selected people who train in groups and finds that they run a mean of 47.1 miles per week. Assume that the population standard deviation for group runners is known to be 4.4 miles per week. She also interviews a random sample of 47 people who train on their own and finds that they run a mean of 48.5 miles per week. Assume that the population standard deviation for people who run by themselves is 1.8 miles per week. Test the claim at the 0.01 level of significance. Let group runners training for marathons be Population 1 and let individual runners training for marathons be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.Suppose the distribution of typing speeds in words per minute for experienced typists can be approximated by a normal curve with a mean of 60 wpm and a standard deviation of 15 wpm. To be considered for a position an applicant needs to have a typing speed of at least 70 wpm. If 62 experienced typists apply for the job, how many of these would be considered based on their typing speed?A marketing researcher wants to estimate the mean amount spent per year ($) on a web site by membership member shoppers. Suppose a random sample of 100 membership member shoppers who recently made a purchase on the web site yielded a mean amount spent of $58and a standard deviation of $55.Complete parts (a) and (b) below. a. Is there evidence that the population mean amount spent per year on the web site by membership member shoppers is different from $51? (Use a 0.01 level of significance.) State the null and alternative hypotheses.
- The gas mileage for a new model of car is normally distributed, with a standard deviation of 3.5 miles per gallon. If only 4% of the cars have a gas mileage of above 20 mpg, what is the mean of the distributionThe president of a college wants to determine the mean number of units that college students take per semester. How large of a sample is required in order to be 99% sure that a sample mean will be off by no more than 1.25 units? An initial study suggested that the standard deviation is approximately 2.1 units.A parenting magazine reports that the average amount of wireless data used by teenagers each month is 5 Gb. For her science fair project, Ella sets out to prove the magazine wrong. She claims that the mean among teenagers in her area is less than reported. Ella collects information from a simple random sample of 25 teenagers at her high school, and calculates a mean of 4.7 Gb per month with a standard deviation of 0.9 Gb per month. Assume that the population distribution is approximately normal. Test Ella's claim at the 0.01 level of significance. Step 3 of 3: Draw a conclusion and interpret the decision. E Tables E Keypad Answer Keyboard Shortcuts We reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance that the average amount of wireless data used by teenagers each month is less than 5 Gb. We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.01 level of significance that the average amount of…
- A study of commuting times reports the travel times to work of a random sample of 1068 employed adults in Chicago. The mean is x = 54.0 minutes and the standard deviation is s = 60.9 minutes. What is the standard error of the mean?Steve believes that his wife's cell phone battery does not last as long as his cell phone battery. On eight different occasions, he measured the length of time his cell phone battery lasted and calculated that the mean was 19.7 hours with a standard deviation of 9.4 hours. He measured the length of time his wife's cell phone battery lasted on twelve different occasions and calculated a mean of 18.9 hours with a standard deviation of 5.3 hours. Assume that the population variances are the same. Let Population 1 be the battery life of Steve's cell phone and Population 2 be the battery life of his wife's cell phone. Step 1 of 2: Construct a 90 % confidence interval for the true difference in mean battery life between Steve's cell phone and his wife's. Round the endpoints of the interval to one decimal place, if necessary.A parenting magazine reports that the average amount of wireless data used by teenagers each month is 10 Gb. For her science fair project, Ella sets out to prove the magazine wrong. She claims that the mean among teenagers in her area is less than reported. Ella collects information from a simple random sample of 16 teenagers at her high school, and calculates a mean of 9.1 Gb per month with a standard deviation of 1.3 Gb per month. Assume that the population distribution is approximately normal. Test Ella's claim at the 0.05 level of significance. Step 3 of 3: Draw a conclusion and interpret the decision.