Question

A plant in the manufacturing sector is concerned about its sulfur dioxide emissions. The weekly sulfur dioxide emissions follow a normal distribution with a mean of 1000 ppm (parts per million) and a standard deviation of 25. Recently, a "cleaner" technology has been adopted. In such a scenario, the CEO would like to investigate whether there has been a significant change in the emissions and has hired you for advice. In other words, the CEO wants to know if the mean level of emissions is different from 1000. Suppose that you are given a sample of data on weekly sulfur dioxide emissions for that plant. The sample size is 50 and x = 1006 ppm. What is the value of the test statistic?a. 2.26b. 1.70c. 4.78d. 2.59

Answer 1

Answer:

The correct option is b) 1.70

Step-by-step explanation:

Consider the provided information.

The weekly sulfur dioxide emissions follow a normal distribution with a mean of 1000 ppm (parts per million) and a standard deviation of 25.

Thus, μ=1000 and σ = 25

The CEO wants to know if the mean level of emissions is different from 1000.

Therefore the null and alternative hypothesis is:

$H_0:\mu =1000$ and $H_a:\mu \neq1000$

The sample size is n = 50 and $\bar x=1006$ ppm.

Now calculate the test statistic by using the formula: $z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}$

Substitute the respective values in the above formula.

$z=\frac{1006-1000}{\frac{25}{\sqrt{50}}}$

$z=\frac{6}{\frac{25}{\sqrt{50}}}=1.697\approx 1.70$

Hence, the correct option is b) 1.70