Question

A sample was taken with 15 observations of soil tests from North Carolina rural yards. We would like to determine whether there is reason to believe the nitrate levels differ from the national average. Our sample average shows 8.19 parts per million (ppm) nitrate concentration with a sample standard deviation of 3.95 ppm. Suppose the national average for rural yard is 6.25 ppm.

Conduct the hypothesis test using a 90% confidence level and answer the following questions. For all numbers, please round to the nearest hundredths.

• What is the standard error?  
• What are the degrees of freedom for this sample?
• What is the T-score?
• Is the p-value less than alpha (type "yes" or "no")?
• What is your conclusion? If you decide they NC yards differ from the US average, type "differ", otherwise type "none".  

Answer 1

Given : Sample size=n=15

Sample mean=8.19

Sample standard deviation=s=3.95

Significance level=0.10

Hypothesized value=6.25

Hypothesis : Vs  

The test is two tailed test.

The standard error is ,

df=degrees of freedom=n-1=15-1=14

The T-score is ,

The p-value is ,

p-value= ; The excel fucntion is , =TINV(1.9021,14,2)

Decision : Here , p-value 0.0779 <

Therefore , reject Ho.

Conclusion : Hence , there is sufficient evidence to support the claim that NC yards differ from the US average.