Question

Show all work including steps used during computations using formulas and diagrams when appropriate. Write a statement explaining the meaning of your computations (conclusions) for all problems- hypothesis tests and confidence intervals.

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))
   

Answer 1

a)

sample mean, xbar = 512.2
sample standard deviation, s = 75.7508
sample size, n = 10
degrees of freedom, df = n - 1 = 9

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.262

ME = tc * s/sqrt(n)
ME = 2.262 * 75.7508/sqrt(10)
ME = 54.185

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (512.2 - 2.262 * 75.7508/sqrt(10) , 512.2 + 2.262 * 75.7508/sqrt(10))
CI = (458.01 , 566.39)

b)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 500
Alternative Hypothesis, Ha: μ > 500

Rejection Region
This is right tailed test, for α = 0.025 and df = 9
Critical value of t is 2.262.
Hence reject H0 if t > 2.262

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (512.2 - 500)/(75.7508/sqrt(10))
t = 0.509

P-value Approach
P-value = 0.3115
As P-value >= 0.025, fail to reject null hypothesis.