Answer:
Step-by-step explanation:
Hello!
1) The objective is to evaluate the depth perception of pilots over the age of fifty.
To do so a sample of n=14 airline pilots over the age of fifty was taken, each of them was asked to judge the distance between two markers placed 20 feet apart, the pilot's error in judging the distance was recorded:
2.9, 2.6, 2.9, 2.6, 2.4, 1.3, 2.3, 2.2, 2.5, 2.3, 2.8, 2.5, 2.7, 2.6
Then the variable of interest is X: error in judging the distance between two markers placed 20 feet apart of one pilot. (feet)
Assuming that this variable has a normal distribution, with a standard deviation of σ= 1.4 feet
The hypothesis is that the mean error in-depth perception for the company's pilots over the age of fifty is greater than 2.1, symbolically: μ > 2.1
H₀: μ ≤ 2.1
H₁: μ > 2.1
α: 0.05
Since the variable has a normal distribution and the population standard deviation is known, the statistic to use for this test is the standard normal:
[tex]Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~~N(0;1)[/tex]
The sample mean is
X[bar]= ∑X/n= 34.90/14= 2.49
[tex]Z_{H_0}= \frac{2.49-2.1}{\frac{1.4}{\sqrt{14} } } = 1.04[/tex]
The p-value for this test is 0.14917
Using the p-value approach, since it is greater than the significance level, the decision is to not reject the null hypothesis.
Then there is no evidence that the mean error in-depth perception for the company's pilots over the age of fifty is greater than 2.1 feet.
2) To construct the 90% confidence interval you have to use the same distribution as before, the formula for the interval under the standard deviation is:
[X[bar] ± [tex]Z_{1-\alpha /2}[/tex] * (δ/√n)]
[tex]Z_{1-\alpha /2}= Z_{0.95}= 1.645[/tex]
[2.49 ± 1.645 * (1.4/√14)]
[1.87; 3.11]
With a 90% confidence, you'd expect that the true mean of the error in-depth perception of the airline's pilots over fifty years old is contained in the interval [1.87; 3.11].
I hope you have a SUPER day!
