Question

Listed below are speeds (mi/h) measured from southbound traffic on I-280 near Cupertino, California (based on data from SigAlert). This simple random sample was obtained at 3:30 pm on a weekday. Use the sample data to construct a 95% confidence interval estimate of the population standard deviation. 62 61 61 57 61 54 59 58 59 69 60 67A.) 4.7 mi/h< <5.6 mi/h

b.) 2.9 mi/h< <6.9 mi/h

c.) 3.1 mi/h< <8.5 mi/h

d.) 1.6 mi/h< <4.9 mi/h

e.) 5.4 mi/h< <9.2 mi/h

Answer 1

Answer:

Option (B) is the correct answer to the following question.

Step-by-step explanation:

Step-1: We have to find the Mean of the series.

The series is Given in the question 62 61 61 57 61 54 59 58 59 69 60 67.

$Mean(\overline{x})=\frac{62+61+61+57+61+54+59+58+59+69+60+67}{12}$ $= 60.67$

Step-2: We have to find the Standard Deviation.

Let Standard Deviation be x.

Formula of Standard Deviation is: $s= \sqrt{\frac{\sum(x_{i}+\overline{x})}{n-1}}$

Put value in formula of Standard Deviation,

$s= \sqrt{\frac{(62+60.67)^{2}+(61+60.67)^{2}+(61+60.67)^{2}+(57+60.67)^{2}+....(67+60.67)^{2}}{n-1}}$ = 40.75

Step-3: Then, we have to find the critical value by chi-square.

$X_{1-\alpha/2}^{2}=3.82$

$X_{1-\alpha/2}^{2}=21.92$

Then, find the confidence interval which is 95%.

$\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{21.92}.(4.075)^2 }\approx2.8868 \\ i.e 2.9$

$\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{3.816}.(4.075)^2 }\approx6.9188 \\ i.e 6.9$