In a clinical trial for an insomnia treatment, a group of 22 subjects had a mean "wake time" of 94.3 minutes after receiving the treatment and a standard deviation of 23.1 minutes.  a) construct a 95% confidence level for the mean.  b) if the subjects had a "wake time" of 100 minutes before

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A: Given data,     P=58%=0.58     E=7%=0.07     z-value at 95% confidence is  zc =1.98     n=?

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A: Solution: Given information: n=36 Sample size  x=2400  Sample mean s= 1200 Sample standard deviation…

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A: Given information: n=31x¯=34s=13 Given confidence level is 98%. Significance level is 2%. degrees of…

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Q: In a survey, 28 people were asked how much they spent on their child's last birthday gift. The…

A: Since population standard deviation is unknown, Use t-distribution to find t-critical value. Find…

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A: In statistical inference, there are two types of estimation,  Point estimation and Interval…

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A: Given : Sample size(n)=2003sample proportion(p^)=0.25confidence level (c)=0.99α=1-c   =1-0.99…

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A: X̄ = 345 n = 20, σ = 20

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A: Given Information: Smokers: n1=150 x1=95 Nonsmokers: n2=250 x2=105 Confidence level is 95%

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A: From the provided information, Sample size (n) = 19 Sample mean (x̄) = $46 Sample standard deviation…

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A: We have given that the 95% confidence interval with mean 32.5, standard deviation 4.5 and sample…

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A: Given, Sample size, n = 43 Sample mean,  = 84 Sample standard deviation, s = 12 Population standard…

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A: From the available information, Sample size, n=30 Sample mean,  Sample standard deviation, s=$19.…

Q: In a survey, 14 people were asked how much they spent on their child's last birthday gift. The…

A: The degrees of freedom is obtained as follows:

Q: In a survey, 31 people were asked how much they spent on their child's last birthday gift. The…

A: Given information Sample mean x̅ = $36 Population standard deviation = $10 Significance level(α) = 1…

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A: Given: n=81x¯=8.2s=9.84 The degrees of freedom is obtained as below: df=n-1=81-1=80 The value of t*…

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A: Givensample size(n)=23sample mean (x) = 31standard deviation (s) = 8α = 1-0.95=0.05α2=0.025degrees…

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Q: In a sample of 95 recommendations, it is found that 56% of them made money. Determine the upper…

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A: Given Sample mean=36 Standard deviation=15 Sample size=24

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A: Given : n= 38.0 , x= 67.0 s= 11.0 At 95.0 % confidence level , z score= 1.96

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A: Given that  Sample size (n)=7  Sample mean = 267  Sample standard deviation (s)= 10  We have to find…

Q: The Associated Press (October 9, 2002) reported that in a survey of 4722 American youngsters aged 6…

A: Obtain the 99% confidence level for the proportion of all American youngsters who are seriously…

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A: In question, We have given the number of favorable cases 150 and sample size 900 then we'll find 95%…