The average student-loan debt is reported to be $25,235. A student believes that the student-loan debt is higher in her area. She takes a random sample of 100 college students in her area and determines the mean student-loan debt is $27,524 and the standard deviation is $6,000. Is there sufficient evidence to support the student's claim at a 5% significance level? Test the claim: Determine the null and alternative hypotheses. Enter correct symbol and value. H0H0: μ=μ= HaHa: μμ? ≠ < > Correct Determine the test statistic. Round to two decimals. t=t= Find the pp-value. Round to 4 decimals. pp-value =

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A: Sample size, n = 31Mean = 48Standard deviation = 8

Q: In a survey, 29 people were asked how much they spent on their child's last birthday gift. The…

A: Givensample size(n)=29mean(x¯)=47standard deviation(s)=2α=1-0.95=0.05α2=0.025Degrees of…

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A: Given,   Sample size = 10 Sample mean = 110000 Population standard deviation = 2000   Since the…

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A: We have to find margin of error

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Q: Suppose you know the population average wage in the manufacturing industry in California is $24.66…

A: Given information: n=30, σ=2.35, μ=24.66α=0.01 Use the excel formula, "=NORM.S.INV(0.01)" for…

Q: In a survey, 27 people were asked how much they spent on their child's last birthday gift. The…

A: Since the population standard deviation is unknown, t distribution can be used for the computation…

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A: From the provided information, Sample size (n) = 49 Sample mean (x̅) = $24.80 Population standard…

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A:

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A: we have given that  sample size =11  ,sample mean= 36500 , standard deviation =5400 and mu=40000

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A:

Q: Decide if the given claim represents the null or alternative hypothesis. a. A governor claimed that…

A:

Q: Test the claim: a. Determine the null and alternative hypotheses. Enter correct symbol and value.…

A: Given  Population mean μ=25235, n=100, sample mean x̄=27524, s=sample standard deviations=6000

Q: at the university spend on groceries each month. From a sample of 29 random students, you find that…

A: Given that  Sample size n =29 Sample mean =871.35 Standard deviation =235.06 NOTE:- According to…

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A: From the provided information, Sample size (n) = 18 Sample mean (x̅) = $59800 Standard deviation (s)…

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A: Given  The results were roughly bell-shaped with a mean of $37 and standard deviation of $3 Sample…

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A: We have given that,  Sample mean (x̄) = $ 44, standard deviation (s) = $ 10, sample size (n) = 30…

Q: engineers in the Philippines

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A:

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A: The summary of statistics is,

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Q: We take a random sample of 70 students in a college in NJ. The mean family income in the sample is…

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Q: (a) Construct and interpret a 95% confidence interval for the mean price paid for coffee in 2015 by…

A: given   n =105       x¯  = 3.42     σ  =0.75 a) confidence interval  limits are    [x¯-   αZ/2 (σ/n)…

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A: Given data,n=31x=562s=17.04df=n-1=30α=0.10t-critical value at α=0.10 and df=30 is tc=1.697

Q: According to the Census Bureau, 3.36 people reside in the typical American household. A sample of 27…

A: Population mean = ? = 3.36 Sample S.D = s =1.25 Sample size = n =27 Sample mean = x̅ = 2.7

Q: a survey, 27 people were asked how much they spent on their child's last birthday gift. The results…

A: Given: Sample size=n=27Mean=x¯=31Sample standard deviation=s=8α=0.01

Q: In a survey, 17 people were asked how much they spent on their child's last birthday gift. The…

A: Given: Sample size,  n=17 Mean,  μ=$30 Standard deviation, s=$10 Confidence level =90% Concept:…

Q: In a survey, 11 people were asked how much they spent on their child's last birthday gift. The…

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A:

Q: In a survey, 10 people were asked how much they spent on their child's last birthday gift. The…

A:

Q: In a survey, 29 people were asked how much they spent on their child's last birthday gift. The…

A: Given that, n=29 mean=$50 S.D=$8

Q: In a data set of 190 ratings, there was a mean rating of 6.82 with a standard deviation of 2.06. Use…

A:

Q: Students who graduated from college in 2010 owed an average of $25,000 in student loans. An…

A: We have given that  Sample size n  = 40 Sample mean = 27500 Population standard deviation = 9120

Q: Suppose a study was conducted of community colleges in California to determine the average cost of…

A: As the population standard deviation is not known, we will use t distribution. The t distribution is…