Question

2 points Save Answer The mean weight of trucks traveling on a particular section of Highway 401 is not known. A provincial highway inspector needs an estimate of the mean. He selects a random sample of 49 trucks passing the weighing station and finds the mean is 15.8 tons, with a standard deviation of the sample of 3 8 tons. What is the 95 percent interval for the population mean? O a 13.2 and 17.6 Ob 16.1 and 18.1 Oc 147 and 16.9 od 10.0 and 20.0 Moving to the next question prevents changes to this answer Question 2 of 35 > e to search ORI e a * 00) 9:16 AM 7/24/2020

Answer 1

The sample mean is 15.8 tons.

(1)

The sample standard deviation (s) is 3.8 tons.

The sample size (n) is 49.

Degrees of freedom = n – 1 = 49 – 1 = 48

At the significance level 0.05 and the degrees of freedom 48, the two-sided critical value obtained from t-table is +/- 2.0106.

The 95% confidence interval for the population mean can be calculated as,

$\begin{array}{c} \bar x \pm \left( {{t^ * } \times \frac{s}{{\sqrt n }}} \right)\\ = 15.8 \pm \left( {2.0106 \times \frac{{3.8}}{{\sqrt {49} }}} \right)\\ = 15.8 \pm 1.09\\ = \left( {14.71,16.89} \right)\\ \approx \left( {14.7,16.9} \right) \end{array}$

Hence, the correct option is c. 14.7 and 16.9