A sample of 145 values is randomly selected from a population with mean, ì, equal to 45 and standard deviation, ó, equal to 23. (Give your answers correct to one decimal place.)
(a) Determine the interval (smallest value to largest value) within which you would expect 99.7% of such sample means to lie.
42.04-47.96
(b) What is the amount of deviation from the mean for a sample mean of 50?
(c) What is the maximum deviation you have allowed for in your answer to part (a)?
(a) To determine the interval within which we would expect 99.7% of sample means to lie, we need to calculate the confidence interval. The formula for the confidence interval is given by:
CI = x̄ ± z(σ/√n)
where CI is the confidence interval, x̄ is the sample mean, z is the z-score corresponding to the desired confidence level, σ is the population standard deviation, and n is the sample size.
In this case, we are given the population mean (ì) as 45 and the population standard deviation (ó) as 23. The sample size (n) is 145.
To find the z-score corresponding to a confidence level of 99.7%, we can use a z-table or a calculator. The z-score for 99.7% confidence level is approximately 2.967.
Substituting the values into the formula, we get:
CI = 45 ± 2.967(23/√145)
Calculating this expression, we find:
CI ≈ 45 ± 2.967(1.913)
CI ≈ 45 ± 5.670
Therefore, the interval within which we would expect 99.7% of sample means to lie is 45 - 5.670 to 45 + 5.670, which simplifies to 39.330 to 50.670. Rounding to one decimal place, the interval is approximately 39.3 to 50.7.
(b) To find the amount of deviation from the mean for a sample mean of 50, we subtract the sample mean (50) from the population mean (45):
Deviation = 50 - 45
Deviation = 5
Therefore, the amount of deviation from the mean for a sample mean of 50 is 5.
(c) The maximum deviation allowed in the answer to part (a) is half the width of the confidence interval. In this case, the confidence interval is from 39.3 to 50.7. To find the width, we subtract the lower bound from the upper bound:
Width = 50.7 - 39.3
Width = 11.4
The maximum deviation allowed is half of the width:
Maximum Deviation = Width/2
Maximum Deviation = 11.4/2
Maximum Deviation ≈ 5.7
Therefore, the maximum deviation allowed in the answer to part (a) is approximately 5.7.