Statistics question confidence intervals ?
I have three problems i have due for homework, i cant seem to figure out. Any help would be helpfull.
1. The average cost of living for family of 4 in a sample of twelve different cities was found to be $65,351 with standard diviation of $7711. What is a 90% confidence inverval with the true mean?
2. A recent study of 900 internet users in Europe found that 20% of the internet users were woman. What is the 95% confidence interval of the true proportion of woman in Europe who uses the internet?
3. The Eagle Ridge sub division claimsd the average price of a home in their subdivision is $112,650 with stanbdard diviation of $6250. A sample of 36 homes for sale in this sub division had an average selling price of 109,850$ using a=.05 does this study supports the claim?
Thank you
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1. ANSWER: 90% Confidence Interval [$56,291, $74,411]
Why???
SMALL-SAMPLE CONFIDENCE INTERVAL FOR A POPLATION MEAN, t-DISTRIBUTION
90% Confidence Interval = x-bar +/- (t-critical value) * σ/SQRT(n)
x-bar = SAMPLE MEAN [$65,351]
σ = STANDARD DEVIATION [$7,711]
n = NUMBER OF SAMPLES [4]
n – 1 = 3 df (DEGREES OF FREEDOM)
t-critical value = (approx) 2.35 from “look-up Table for “two-sided interval” df = 3
90% Confidence Interval: $65351+/- 2.35 * $7,711 / SQRT(4) = [$56,291, $74,411]
That is; with a confidence interval of approximately 90% the “true mean” (μ) is within the interval of [$56,291, $74,411] and that the sample mean (x-bar) which is estimate of “true mean” (μ) is $801.75.
2. ANSWER: 95% Confidence Interval 0.178, 0.222] (17.8% , 22.2%)
Why???
POPULATION PROPORTION, CONFIDENCE INTERVAL, NORMAL DISTRIBUTION
95% CONFIDENCE INTERVAL p +/- (z critical value) * SQRT[p * (1 - p)/n]
p = POPULATION PROPORTION [0.20] (20%)
z critical value [1.645]
n = SAMPLE SIZE [900]
95% CONFIDENCE INTERVAL = 0.20+/- 1.645 * SQRT [ 0.20 * (1 - 0.20)/900] = [0.178, 0.222] (17.8% , 22.2%)
CONCLUSION: 95% CONFIDENCE INTERVAL of true POPULATION PROPORTION =[0.178, 0.222] (17.8% , 22.2%)
3. ANSWER: To a significance value α = 0.05, Average price of a home ≠ $112,650
Why??
SINGLE SAMPLE TEST, “two-tailed test” 7 – Step Procedure for t Distributions
1. Parameter of interest: “μ” = population mean for average price of a home
2. Null hypothesis Ho: μ = $112,650
3. Alternative hypothesis Ha: μ ≠ $112,650
4. Test statistic formula: t = (x-bar – μ)/(s/SQRT(n))
x-bar = estimate of the Population Mean (statistical mean of the sample) [$109,850]
n=number of individuals in the sample [36]
s=standard deviation [$6,250]
μ=Population Mean [$112,650]
5. Computation of Test statistic formula t = (x-bar – μ)/(s/SQRT(n))
t = ($112,650 – $109,850)/($6,250/SQRT(36)) = 2.688
6. Determination of the P-value: test based on n -1 = 35 df (degrees of freedom). Table “look-up” value shows area under 35 df curve to right of t = 2.688 is (approx.) 0.005
7. Conclusion: For significance value α = 0.05/2 “two-tailed”), above shows P-value <= α, [0.005 <= 0.05/2] so there is compelling reason to reject Null hypothesis Ho: μ = $112,650 (average price of a home) and accept Alternative hypothesis Ha: μ ≠ $112,650.