Question 1065358: #1
You want to determine a confidence interval with a confidence level of 95%
And 99% for the proportion of housewives who buy only once a week.
If known
That in a simple random sample of 400 housewives only 180 of said to buy one
Once a week.
#2
A sample of 25 students from a Faculty has been obtained to estimate the
Average grade of student records in the Faculty. It is known by other courses.
That the standard deviation of the scores in said Faculty is of 2.01 points. If the average Of the sample was 4.9. Calculates:
1. Confidence interval at 90%.
2. 95% Confidence Interval
3. Confidence interval at 99%
#3
We want to obtain a confidence interval for the value of average sales
Per hour that occur in a kiosk. For this, we perform a sample consisting of choosing At random the sales that were made during 1000 different hours; Shows whose results
Were: average sales per hour 4000 pesos, and a standard deviation of 63,245 pesos. Get Said interval with a confidence level of 95%.
#4
The average height of a random sample of 400 people from a City is 1.75 m. It is known that the height of the people of that city is a aleatory variable that follows a normal distribution with variance σ2 = 0.16 m2.
(A) Constructs an interval of 95% confidence for the mean of the height of the population.
(B) What would be the minimum sample size necessary to be able to be said to
be True height of the statures is less than 2 cm from the sample mean, with a 90% confidence level?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 1. One sample proportion, p=0.45, 1-p=0.55
interval width for 95% is z(0.975)* sqrt ((p*(1-p))/n)=1.96* sqrt (0.2475/400)=0.0249*1.96=0.0488
(0.4012, 0.4988).
For 99%, z=2.576 and interval width is 0.0641 for (0.3859, 0.5141)
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2. The 95% CI is x bar +/- t (df=24, 0.975)*2.01/sqrt (25)
interval width is 2.064*2.01/5=0.83
95% interval is (4.07, 5.73)
90% interval uses 1.711 for t and has interval width of 0.70; (4.2, 5.6)
99% interval uses 2.797 for t and has interval width of 1.12; (3.78, 6.02)
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3. This doesn't appear to work with mean of 4000 and sd of 63,245; if such an sd is in fact the case, the distribution is not normal and can't use that. I think the 63.245 would likely be too small a sd.
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4.Mean is 1.75 m and sigma is 0.4 m (sqrt of 0.16 m^2)
95% CI interval width is is 1.96*0.4/20=0.0392
(1.71, 1.79)
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Need an interval width of 2 with z=1.645, sd 0.4 and n unknown
2=1.645*0.4/ sqrt (n)
sqrt (n)=1.645*0.4/0.02=32.9
square both sides, and n=1082.41=1083.
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