Moviegoers 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 47 moviegoers from a large university revealed that the average number of movies seen per person was 9.1. 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? Part: 0/ 5 Part 1 of 5 State the hypotheses and identify the claim with the correct hypothesis. (Choose one) ▼ ロ<ロ H : ロ=ロ (Choose one) v ロロ This hypothesis test is a (Choose one) v test.

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A: Given, Population mean=39385 Population standard deviation=5975 sample mean=41680 sample size = 50…

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A: Hi! Thanks for posting the question. Since there are more than three sub-part questions, we will…

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A: Given information- Population mean, μ = $20,000 Sample size, n = 22 Sample mean, x-bar = $20,640…

Q: A medical investigation claims that the average number of infections per week at a hospital in Town…

A: Solution-: Given: x¯=17.7,μ0=16.3,n=40,s=1.8,α=0.05 Claim: The average number of infections per week…

Q: ​Full-time college students report spending a mean of 29 hours per week on academic​ activities,…

A: From the provided information, Mean (µ) = 29 hours Standard deviation (σ) = 4 X~N (29, 4) Here X be…

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A: GivenMean(x)=7.1standard deviation(s)=2.2sample size(n)=16α=0.01

Q: The average "moviegoer" sees 8.5 movies a year. A moviegoer is defined as a person who sees at least…

A: We have given that  Sample size n = 49  Sample mean = 9.3 Population standard deviation = 3.2

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A: Given For the first question μ=50000σ=3500   For the second question  μ=50σ=6

Q: In high school X, approximately 9 percent of the students saw a certain movie on opening night. From…

A: In high school X, approximately 9 percent of the students saw a certain movie on opening night. From…

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A: Calories in a mid-sized hamburger: 515 506 501 498 497 506 458 478 463 514 Sample size = n =…

Q: Executives of a supermarket chain are interested in the amount of time that customers spend in the…

A: The provided information is:

Q: At the 0.10 level of significance, can it be concluded that this represents a difference from the…

A: Given that Sample size n = 50 Sample mean = 9.2 Population SD = 3.2 Level of significance = 0.10

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A: Null and alternative hypotheses:Let σ is the standard deviation of machine dispenses amount.Null…

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A:

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A: The population mean is denoted by μμ. The population standard deviation is denoted by σσ. Sample…

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A:

Q: According to previous studies, the mean distance each visitor in Greenspan National Park hikes…

A: From the above data  The test statistic  Z=0.339 P value is 0.367 Level of significance =0.10

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A: According to the provided information, The hypothesis can be defined as:

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A: a. Null and Alternative hypotheses: From the given information, the objective is to test whether or…

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A: Solution: State the hypotheses. Null hypothesis: That is, the population standard deviation of TV…

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A: Given, n =  μ = 13252x¯ = 15560s = 3500α = 0.01

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A: Given problem Given that   Credit Card Debt  It has been reported that the average credit card…

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A: Given , Sample size = n = 32 Sample standard deviation = s = 3.2 Claim : The population standard…

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Q: In the field of accountancy, times spent studying by students in the week before professional exams…

A: 1. a)  Probability that sample mean exceeds population mean by more than 2 hours is given by :…

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A: Given that  μ=57337σ=7500 A) The probability that a randomly selected teacher makes less than…

Q: The average "moviegoer" sees 8.5 movies a year. A moviegoer is defined as a person who sees at least…

A: Given Average =9.9 Sample SD=3.2 n=42 μ=8.5

Q: . According to Dietary Goals for the US (1977), high sodium intake may be related to ulcers, stomach…

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