A sample of 900 members has a mean 3.4 cm and standard deviation 2.61 cm. Test whether the sample is from a large population of mean 3.25 cm and standard deviation 2.61 cm. If the
population is normal and its mean is unknown, find the 95% confidence interval for population
mean.
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N_s = sample size = 900. μ_s = mean of the sample = 3.4 cm
σ_s = Standard deviation of the sample = 2.61 cm
Population mean μ₀ =3.25
Population standard deviation = σ₀ = 2.61 cm
student's t = (μ_s - μ₀) / [σ_s / √N_s ]
t = (3.4 - 3.25) * √900 / 2.61 = 1.7241
Find the probability that -1.7241 <= t <= 1.7241 from the students t distribution table or from a website that calculates these. Here the degrees of freedom (d.f) are 899 = N_s - 1. Sample size is 900.
The probability we get is 0.915. Hence, we can say that the sample belongs to the population with a probability 0.915 or a confidence level of 91.5%.
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The population is normal. we don't know its mean. We know sample size, mean and standard deviation. We are given confidence level 95% = 0.95
Normal distribution variable:
Z = (μ_s - μ₀) / (σ_s/√N_s)
Here σ_s = 2.61 cm, N_s = 900
We consult the Standard Normal Distribution tables. We find z such that
P(0 <= Z <= z ) = 0.95/2 = 0.475,
or from the cumulative standard distribution table we find z such that :
P(-∞ < Z <= z ) = 0.5 + 0.95/2 = 0.975
Either way, we get the value z of variable Z for which the probability
P( -z <= Z <= z) = 95% = 0.95 is z = 1.96
-1.96 <= Z <= 1.96
-1.96 * σ_s <= √N_s * (μ_s - μ₀) <= 1.96 * σ_s
- 0.1705 <= (μ_s - μ₀) <= 0.1705 cm
As, μ_s = 3.4 cm :
3.4 - 0.1705 <= μ₀ <= 3.4+0.1705 cm
ie., range [3.23 cm , 3.57 ]
That will be the range of the population mean that we will estimate from the available sample data.
σ_s = Standard deviation of the sample = 2.61 cm
Population mean μ₀ =3.25
Population standard deviation = σ₀ = 2.61 cm
student's t = (μ_s - μ₀) / [σ_s / √N_s ]
t = (3.4 - 3.25) * √900 / 2.61 = 1.7241
Find the probability that -1.7241 <= t <= 1.7241 from the students t distribution table or from a website that calculates these. Here the degrees of freedom (d.f) are 899 = N_s - 1. Sample size is 900.
The probability we get is 0.915. Hence, we can say that the sample belongs to the population with a probability 0.915 or a confidence level of 91.5%.
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The population is normal. we don't know its mean. We know sample size, mean and standard deviation. We are given confidence level 95% = 0.95
Normal distribution variable:
Z = (μ_s - μ₀) / (σ_s/√N_s)
Here σ_s = 2.61 cm, N_s = 900
We consult the Standard Normal Distribution tables. We find z such that
P(0 <= Z <= z ) = 0.95/2 = 0.475,
or from the cumulative standard distribution table we find z such that :
P(-∞ < Z <= z ) = 0.5 + 0.95/2 = 0.975
Either way, we get the value z of variable Z for which the probability
P( -z <= Z <= z) = 95% = 0.95 is z = 1.96
-1.96 <= Z <= 1.96
-1.96 * σ_s <= √N_s * (μ_s - μ₀) <= 1.96 * σ_s
- 0.1705 <= (μ_s - μ₀) <= 0.1705 cm
As, μ_s = 3.4 cm :
3.4 - 0.1705 <= μ₀ <= 3.4+0.1705 cm
ie., range [3.23 cm , 3.57 ]
That will be the range of the population mean that we will estimate from the available sample data.
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Answer:
Solution:
(i) Given:
Sample size n = 900, Sample mean = 3.4 cm, Sample SD σ = 2.61 cm
Population mean μ= 3.25 cm, Population SD σ = 2.61 cm
Null Hypothesis H0 : μ = 3.25 cm (the sample has been drawn from the population mean
μ = 3.25 cm and SD σ = 2.61 cm)
Alternative Hypothesis H1 : μ ≠ 3.25 cm (two tail) i.e., the sample has not been drawn from the population mean μ = 3.25 cm and SD σ = 2.61 cm.
The level of significance α = 5% = 0.05
Teststatistic:
∴ Z = 1.724
Thus the calculated and the significant value or table value Zα/2 = 1.96
Comparing the calculated and table values, Z < Zα/2 i.e., 1.724 < 1.96
Step-by-step explanation:
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