Question

8. The national standard for salinity (salt content) in drinking water is 500 parts per million. A government inspector tests the drinking water at 10 locations around a certain city and finds the following levels of salinity (in parts per million of course):

532

560

506

452

425

531

582

376

529

629

1. Make a 95% confidence interval for the mean salinity level in this city.
2. Is this city’s salinity level above the national standard? (Use α = 0.025 and the result from part (a))

please explain part b in detail. don't copy and paste what is already uploaded

Answer 1

a)
From the given data,
mean = 512.20
s = 75.7508 , n =10

The t value at 95% confidence interval is,

alpha = 1 - 0.95 = 0.05
alpha/2 = 0.05/2 = 0.025
t(alpha/2,df) = t(0.025,9) = 2.262

CI = mean +/- t *(s/sqrt(n))
= 512.20 +/- 2.262*(75.7508/sqrt(10))
= (458.0149 , 566.3851)

b)

H0 : mu = 500
Ha: mu not =500

test statistics:

t = (xbar -mu)/9s/sqrt(n))
= ( 512.20 - 500)/(75.7508/sqrt(10))
= 0.51

p value = 0.6228

As p value > 0.025 Do not reject H0

The city’s salinity level is not above the national
standard

As the in part a) the confidence interval contain hypothesised value 500 so, do not reject H0