A chemical plant is required to maintain ambient sulfur levels in the working environment atmosphere at an average level of no more than 12.50. The results of 15 randomly timed measurements of the sulfur level produced a sample mean of Y = 14.82 and a sample standard deviation of s = 2.91. Required:
Is there evidence that the chemical plant is in violation of the working code?

Since the calculated value of z= 3.101 falls in the critical region it is concluded that average level of the chemical is greater than 12.50. The null hypothesis is rejected. There is evidence that the chemical plant is in violation of the working code.

Step-by-step explanation:

Here

The population mean = u= 12.50

The sample mean = x`= 14.83

The sample size= n= 15

Sample standard deviation= 2.91

Let the null and alternate hypotheses be

H0  : u ≤ 12.50 against the claim that Ha: u > 12.50

Applying z- test

z= x`-u/s/√n

Z= 14.83-12.5/ 2.91/√15

Z= 2.33/ 0.751

z= 3.101

The significance level is chosen to be 0.05

The critical value of z at 0.05 > 1.645 for one tailed test

Since the calculated value of z= 3.101 falls in the critical region it is concluded that average level of the chemical is greater than 12.50. The null hypothesis is rejected. There is evidence that the chemical plant is in violation of the working code.

30 miles

step-by-step explanation:

we know that 60 seconds equals 1 minute and 60 minutes 1 hour

44 ft       60 sec     60 min     158400   ft

  *   *   =

1 sec       1 min       1 hour         hour

we know that there are 5280 ft per mile

158400   ft       1 mile         30 miles

  *   =

1   hour               5280 ft     1 hour

so in 1 hour   it can travel 30 miles

sure i'm up for it

step-by-step explanation: