In the 2000 census,the so-called << long form>>received by one of every six households contained 52 questions, ranging from your occupation and income all the way to wether you had abath tub.According to the us.census Bureau,the mean completion time for the long form is 38 minutes. Assuming standard deviation of 5 minutes and a simple random sample of 50 person who filled out the form, what is the probability that theiraverage time for completion of the form wasare than 45 minutes?
We can use the central limit theorem to approximate the sampling distribution of the sample mean. Since the sample size is large (n=50) and the population standard deviation is known, we can use the z-score formula:
z = (x̄ - μ) / (σ / √n)
where x̄ is the sample mean, μ is the population mean (38 minutes), σ is the population standard deviation (5 minutes), and n is the sample size (50).
We want to find the probability that the sample mean is greater than 45 minutes:
P(x̄ > 45) = P(z > (45 - 38) / (5 / √50))
P(z > 4.71) = 1 - P(z < 4.71)
Using a standard normal table or calculator, we find that P(z < 4.71) = 0.9999999984 (rounded to 10 decimal places).
Therefore, P(z > 4.71) = 1 - P(z < 4.71) = 1 - 0.9999999984 = 1.6 x 10^-9 (rounded to scientific notation).
The probability that the average completion time for the sample of 50 people is more than 45 minutes is extremely small, less than 0.0000002%. This suggests that it is highly unlikely that the sample mean would be so different from the population mean by chance alone, and we might question whether the sample is truly representative of the population.
The folowing data show the number of hours perday 12 adults spent in front of screens watching TV content and those selected from normal distribution.
2 5 4 4 6 7
4 2 3 1 2 3
Construct a 95% confidence interval to estimate the average number of hours perday adults spend watching Television.
We can use a t-distribution to construct a confidence interval for the true population mean. Since the sample size is small (n=12) and the population standard deviation is unknown, we can use the sample standard deviation as an estimate and the t-distribution formula:
t = (x̄ - μ) / (s / √n)
where x̄ is the sample mean, μ is the population mean, s is the sample standard deviation, and n is the sample size.
First, we calculate the sample mean and standard deviation:
x̄ = (2+5+4+4+6+7+4+2+3+1+2+3) / 12 = 3.5 hrs/day
s = sqrt( Σ(xi - x̄)^2 / (n-1) ) = 1.875 hrs/day
Next, we need to find the t-value with 11 degrees of freedom (n-1) and a 95% confidence level. Using a t-distribution table or calculator, we find that the t-value is 2.201.
Finally, we can construct the confidence interval:
CI = x̄ ± t(s / √n)
CI = 3.5 ± 2.201(1.875 / √12)
CI = 3.5 ± 1.288
The 95% confidence interval for the average number of hours per day adults spend watching television is (2.212, 4.788) hours/day.
We are 95% confident that the true population mean falls within this interval. This means that if we were to repeat this study many times and construct 95% confidence intervals for each sample, about 95% of them would contain the true population mean.
To solve this problem, we will use the Central Limit Theorem. According to the theorem, the sampling distribution of the sample mean approaches a normal distribution as the sample size increases, regardless of the shape of the population distribution.
In this case, we are given the mean completion time for the long form as 38 minutes with a standard deviation of 5 minutes. We are also given a sample size of 50.
First, we calculate the standard error of the mean (SEM) using the formula:
SEM = standard deviation / sqrt(sample size)
SEM = 5 / sqrt(50)
SEM ≈ 0.707
Next, we convert the desired completion time of 45 minutes into a z-score using the formula:
z-score = (x - mean) / SEM
z-score = (45 - 38) / 0.707
z-score ≈ 9.899
With the z-score calculated, we can now find the probability using a standard normal distribution table or calculator. However, since the z-score of 9.899 is very large, the probability is essentially zero.
In conclusion, the probability that the average completion time for the sample of 50 people was more than 45 minutes is practically zero.