Answer to Question #124521 in Statistics and Probability for Jessica

Question

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 554 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

a) The provided sample mean is $\bar{X}=2770$ 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 $\alpha=0.05.$

Based on the provided information, the critical t-value for $\alpha=0.05$ and $df=19$ degrees of freedom is  $t_c=2.093.$

The 95% confidence for the population mean $\mu$ is computed using the following expression

$$CI=(\bar{X}-\dfrac{t_c\times s}{\sqrt{n}},\bar{X}+\dfrac{t_c\times s}{\sqrt{n}})=$$

$$=(2770-\dfrac{2.093\times 554}{\sqrt{20}},2770+\dfrac{2.093\times 554}{\sqrt{20}})=$$

$$=(2511, 3029)$$

$$2511\leq\mu\leq3029$$

b) The critical t-value for $\alpha=0.05$ and $df=19$ degrees of freedom (left-tailed) is $t_c=-1.729.$

$$\bar{X}+\dfrac{t_c\times s}{\sqrt{n}}=2770-\dfrac{1.729\times 554}{\sqrt{20}}=2556$$

$$\mu\leq2556$$

c) The critical t-value for $\alpha=0.05$ and $df=19$ degrees of freedom (right-tailed) is  $t_c=1.729.$

$$\bar{X}+\dfrac{t_c\times s}{\sqrt{n}}=2770+\dfrac{1.729\times 554}{\sqrt{20}}=2984$$

$$\mu\geq2984$$