Answer to Question #124521 in Statistics and Probability for Jessica

Question #124521
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.
1
Expert's answer
2020-07-02T19:34:17-0400

a) The provided sample mean is Xˉ=2770\bar{X}=2770 and the sample standard deviation is s=554.s=554.

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

The number of degrees of freedom are df=201=19,df=20-1=19, and the significance level is α=0.05.\alpha=0.05.

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

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


CI=(Xˉtc×sn,Xˉ+tc×sn)=CI=(\bar{X}-\dfrac{t_c\times s}{\sqrt{n}},\bar{X}+\dfrac{t_c\times s}{\sqrt{n}})=

=(27702.093×55420,2770+2.093×55420)==(2770-\dfrac{2.093\times 554}{\sqrt{20}},2770+\dfrac{2.093\times 554}{\sqrt{20}})=

=(2511,3029)=(2511, 3029)

2511μ30292511\leq\mu\leq3029

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


Xˉ+tc×sn=27701.729×55420=2556\bar{X}+\dfrac{t_c\times s}{\sqrt{n}}=2770-\dfrac{1.729\times 554}{\sqrt{20}}=2556

μ2556\mu\leq2556


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


Xˉ+tc×sn=2770+1.729×55420=2984\bar{X}+\dfrac{t_c\times s}{\sqrt{n}}=2770+\dfrac{1.729\times 554}{\sqrt{20}}=2984

μ2984\mu\geq2984



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