Respuesta :
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:
[tex]H_0:\mu =1000[/tex] and [tex]H_a:\mu \neq1000[/tex]
The sample size is n = 50 and [tex]\bar x=1006[/tex] ppm.
Now calculate the test statistic by using the formula: [tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
Substitute the respective values in the above formula.
[tex]z=\frac{1006-1000}{\frac{25}{\sqrt{50}}}[/tex]
[tex]z=\frac{6}{\frac{25}{\sqrt{50}}}=1.697\approx 1.70[/tex]
Hence, the correct option is b) 1.70
