In a sample of 100 steel wires the average breaking strength is 50 kN, with a standard deviation of 2 kN.
a. Find a 95% confidence interval for the mean breaking strength of this type of wire.
b. An engineer claims that the mean breaking strength is between 49.7 kN and 50.3 kN. With what level of
confidence can this statement be made?
c. How many wire must be sampled so that a 95% confidence interval specifies the mean breaking strength
to within ±0. 3 kN?
oten
To find the answers to these questions, we will use the concept of confidence intervals and the formula for calculating the margin of error.
a. Find a 95% confidence interval for the mean breaking strength of this type of wire:
To find the confidence interval, we need to consider the sample mean, standard deviation, sample size, and the level of confidence.
The formula for calculating the confidence interval is:
Confidence interval = Sample mean ± (Z * (Standard deviation / √Sample size))
Since we are interested in a 95% confidence interval, the Z-value (critical value) is 1.96. This value represents the number of standard deviations needed to include 95% of the data.
Substituting the given values into the formula:
Confidence interval = 50 kN ± (1.96 * (2 kN / √100))
Simplifying the formula:
Confidence interval = 50 kN ± (1.96 * 0.2 kN)
So, the 95% confidence interval for the mean breaking strength is: (49.608 kN, 50.392 kN)
b. An engineer claims that the mean breaking strength is between 49.7 kN and 50.3 kN. With what level of confidence can this statement be made?
To determine the level of confidence, we need to see if the claimed interval falls within the 95% confidence interval we calculated in part a.
If the claimed interval (49.7 kN, 50.3 kN) falls within the 95% confidence interval (49.608 kN, 50.392 kN), it means that the statement can be made with at least 95% confidence.
In this case, the engineer's claim does fall within the 95% confidence interval, so the statement can be made with 95% confidence.
c. How many wires must be sampled so that a 95% confidence interval specifies the mean breaking strength to within ±0.3 kN?
To determine the sample size needed, we can use the formula for calculating the margin of error:
Margin of error = Z * (Standard deviation / √Sample size)
We want the margin of error to be ±0.3 kN. Since the Z-value for a 95% confidence interval is 1.96 (as calculated in part a), we can plug in these values and solve for the sample size.
0.3 kN = 1.96 * (2 kN / √Sample size)
Simplifying the formula:
√Sample size = (1.96 * 2 kN) / 0.3 kN
Squaring both sides of the equation:
Sample size = [(1.96 * 2 kN) / 0.3 kN]^2
Evaluating the expression:
Sample size = 67.07
To ensure an accurate and reliable estimate with a 95% confidence interval and a margin of error of ±0.3 kN, a sample size of at least 68 wires should be taken.