Answer to Question #124518 in Statistics and Probability for jse

Question

2. The number of steps to cover the same distance was measured for 20 randomly chosen people. The sample mean is 2770 steps and the sample standard deviation is steps. Assume distribution to be normal.

Round your answers to the nearest integer (e.g. 9876).


a) Construct a 95% two-sided confidence interval on the mean.
b) Construct a 95% lower confidence bound on the mean.
c) Construct a 95% upper confidence bound on the mean.

Answer 1

Given : sample mean = 2770, sample sd = 554, n= 20

To find : confidence interval for 95%since population sd is unknown we use confidence interval for t.

Formula: Lower bound =$\overline{x}-\frac{tc*s}{\sqrt{n}}$

Upper bound =$\overline{x}+\frac{tc*s}{\sqrt{n}}$

Two tailed confidence interval =$[\overline{x}-\frac{tc*s}{\sqrt{n}}\leq\mu\leq\overline{x}+\frac{tc*s}{\sqrt{n}}]$

The provided sample mean is  = 2770 and the sample standard deviation is s = 554

The size of the sample is n = 20 and the required confidence level is 95%.

The number of degrees of freedom are df = 20 - 1 = 19, and the significance level is α=0.05.

Using t table tc=2.093.

solution :

Part a)

Two sided confidence interval on the mean:

$[2770-\frac{2.093*554}{\sqrt{20}}\leq \mu\leq 2770+\frac{2.093*554}{\sqrt{20}}]$

=(2770−259.28,2770+259.28)

= (2510.72, 3029.28)

=(2510.72,3029.28)

part b) Lower bound =$2770-\frac{2.093*554}{\sqrt{20}}$

=2510.72

Part c) Upper bound =$2770+\frac{2.093*554}{\sqrt{20}}$

=3029.28