(25 pts, 5 pts each) The ounces of coffee that a machine dispenses in 5-ounce cups have normally distributed. Suppose a sample of size 6 was taken from this machine, as shown as follows:
5.05 5.03 5.10 5.02 4.98 5.00
(for all calculations, keep at least six decimal places)
(1) Find the point estimations for the population mean , variance 2 and standard deviation .
Ans.
(2) Find the 95% confidence interval for . (t variable)
Ans.
(3) Without a computation, will the 90% confidence interval for longer or shorter than the confidence interval you obtained in (2)? Why?
Ans.
(4) Find the 95% confidence interval for 2 and , respectively. (2 variable)
Ans.
(5) If a sample of size 30 was taken, what statistic you would use in (2)? Why?
Ans.
We do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though.
However, I will start you out.
Find the mean first = sum of scores/number of scores.
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
(1) Point estimations:
To find the point estimations for the population mean μ, variance σ^2, and standard deviation σ, we need to calculate the sample mean (x̄) and sample variance (s^2) first.
Sample mean (x̄) = (5.05 + 5.03 + 5.10 + 5.02 + 4.98 + 5.00) / 6 = 5.033333...
Sample variance (s^2) = [(5.05 - 5.033333...)^2 + (5.03 - 5.033333...)^2 + (5.10 - 5.033333...)^2 + (5.02 - 5.033333...)^2 + (4.98 - 5.033333...)^2 + (5.00 - 5.033333...)^2] / (6 - 1) = 0.001666...
Point estimations:
Population mean (μ) = Sample mean (x̄) = 5.033333...
Variance (σ^2) = Sample variance (s^2) = 0.001666...
Standard deviation (σ) = √(Variance) = √(0.001666...) = 0.040822...
(2) 95% Confidence interval for μ (t variable):
To find the confidence interval, we need to calculate the standard error (SE).
Standard error (SE) = Standard deviation (σ) / √(sample size) = 0.040822... / √(6) = 0.016693...
Using the t-distribution table or a t-distribution calculator with α = 0.05 and (n-1) degrees of freedom (df = 6-1 = 5), we can find the t-value. For a 95% confidence interval, the critical t-value is approximately 2.571.
Confidence interval for μ = Sample mean (x̄) ± (t-value * SE)
Confidence interval for μ = 5.033333... ± (2.571 * 0.016693...)
Confidence interval for μ = 5.033333... ± 0.042887...
So, the 95% confidence interval for μ is (4.990447..., 5.076220...).
(3) Without a computation, the 90% confidence interval for μ will be shorter than the 95% confidence interval obtained in (2). This is because as the level of confidence decreases, the margin of error also decreases, resulting in a narrower interval.
(4) 95% Confidence interval for σ^2 and σ (χ^2 variable):
To find the confidence interval, we need to calculate the lower and upper critical χ^2-values and then convert them to the lower and upper limits of the confidence interval for the population variance (σ^2) and standard deviation (σ).
Lower critical χ^2-value = (sample size - 1) * (sample variance / χ^2 value at α/2 with (n-1) degrees of freedom)
Lower critical χ^2-value = 5 * (0.001666...) / χ^2 value at 0.025 with 5 degrees of freedom = 0.007248...
Upper critical χ^2-value = (sample size - 1) * (sample variance / χ^2 value at 1 - α/2 with (n-1) degrees of freedom)
Upper critical χ^2-value = 5 * (0.001666...) / χ^2 value at 0.975 with 5 degrees of freedom = 0.074318...
Confidence interval for σ^2 = [(n-1) * sample variance] / upper critical χ^2-value to [(n-1) * sample variance] / lower critical χ^2-value
Confidence interval for σ^2 = [5 * 0.001666...] / 0.074318... to [5 * 0.001666...] / 0.007248...
Confidence interval for σ^2 = 0.111161... to 0.288358...
Confidence interval for σ = √Confidence interval for σ^2
Confidence interval for σ = √(0.111161...) to √(0.288358...)
Confidence interval for σ = 0.333551... to 0.537067...
So, the 95% confidence interval for σ^2 is (0.111161..., 0.288358...) and for σ is (0.333551..., 0.537067...).
(5) If a sample of size 30 was taken, we would still use the t-distribution in (2) to find the confidence interval for μ. This is because the population distribution is normally distributed and the sample size is small (less than 30). For larger sample sizes (generally greater than 30), the t-distribution approaches the standard normal distribution.
(1) To find the point estimations for the population mean μ, variance σ^2, and standard deviation σ, we can use the sample data.
The population mean can be estimated by calculating the sample mean (x̄) which is the average value of the observations in the sample. To get x̄, sum up all the values and divide by the sample size (n). Therefore,
x̄ = (5.05 + 5.03 + 5.10 + 5.02 + 4.98 + 5.00) / 6 = 5.033333...
So, the point estimation for the population mean μ is approximately 5.033333.
The variance σ^2 can be estimated by calculating the sample variance (s^2) which measures the spread or variability of the data. To find s^2, subtract the sample mean from each observation, square the differences, sum them up, and divide by (n-1). Therefore,
s^2 = [(5.05-5.033333)^2 + (5.03-5.033333)^2 + (5.10-5.033333)^2 + (5.02-5.033333)^2 + (4.98-5.033333)^2 + (5.00-5.033333)^2] / (6-1)
= 0.000350...
So, the point estimation for the population variance σ^2 is approximately 0.000350.
Finally, the standard deviation σ can be estimated by taking the square root of the variance. Therefore,
σ = √(0.000350...)
So, the point estimation for the population standard deviation σ is approximately 0.018708....
(2) To find the 95% confidence interval for μ using a t distribution, we need to know the critical value for a 95% confidence level with our sample size (n=6).
Since the sample size is small (n<30), we use the t-distribution to calculate the critical value.
To find the critical value, we need to know the degrees of freedom. The degrees of freedom for estimating the population mean with a sample size of n is n-1. In our case, the degrees of freedom are 6-1 = 5.
Using a t-table or a statistical calculator, the critical value for a 95% confidence level with 5 degrees of freedom is approximately t = 2.571.
The 95% confidence interval for the population mean μ can be calculated using the formula:
CI = x̄ ± (t * (s/√n))
where x̄ is the sample mean, t is the critical value, s is the sample standard deviation, and n is the sample size.
Plugging in the values, we have:
CI = 5.033333 ± (2.571 * (0.018708/√6))
Calculating this expression gives us the range of the confidence interval.
(3) Without computation, we can infer that the 90% confidence interval for μ will be shorter than the 95% confidence interval obtained in (2). This is because as we decrease the confidence level, we need less certainty in our estimation, which allows for a narrower range.
(4) To find the 95% confidence interval for σ^2 and σ, we need to use the chi-squared (χ^2) distribution.
The lower and upper bounds of the confidence interval for σ^2 can be calculated using the chi-squared distribution with (n-1) degrees of freedom. The chi-squared distribution has a lower and upper critical value associated with a certain confidence level.
Using a chi-squared table or a statistical calculator, we can find the lower and upper critical values for a 95% confidence level with (n-1) degrees of freedom.
The 95% confidence interval for σ^2 can be calculated using the formula:
CI = [(n-1)*s^2 / χ^2_upper, α/2, (n-1)] , [(n-1)*s^2 / χ^2_lower, 1-α/2, (n-1)]
where s^2 is the sample variance, χ^2_upper is the upper critical value, χ^2_lower is the lower critical value, and α is (1 - confidence level).
To find the 95% confidence interval for σ, we just need to take the square root of the lower and upper bounds of the confidence interval for σ^2.
(5) If a sample of size 30 was taken, we would use the z statistic in (2) instead of the t statistic. This is because when the sample size is large (n≥30), the sampling distribution becomes approximately normal and can be approximated using the standard normal distribution (z distribution). In this case, we can use the z critical value instead of the t critical value to find the confidence interval.