Question

The average dog can run 0.4 miles before having to stop, with a standard deviation of 0.09 miles, and the distribution of these distances is roughly normal. We take a sample of 16 dogs and prescribe a daily training routine. After completing training, the dogs had a mean run distance of 0.45 miles. We will use a 0.10 significance level to see whether the training can increase mean run distance for all dogs.

a. What is the population?

b. What is the sample?

c. What is an individual?

d. What is the variable?

e. Is the variable quantitative or qualitative?

f. State the null hypothesis.

g. State the alternate hypothesis

. h. Give the tail type.

i. Compute the test statistic.

j. Compute the observed significance level (?-value).

k. Make a statistical conclusion

. l. State your conclusion in plain English.

Answer 1

a) the popuplation is from where the sampe is taken , and is normally distributed , so here ,

mu = 0.45 , is the population parameter

b) sample is denoted by , x , here , x = 0.4 is the sample parameter

c) in this scenerio , an individual is the dog , it is the object or subject in this problem .

d) the variable in this case is defined as how many miles an average can run .

e) the variable is quantitative because data is about numeric variables in miles .

f) the null hypothesis :

H0 : mu <= 0.45 ,

g) the alternate hypothesis :

Ha : mu > 0.45 , where mu is the mean run distance for all dogs.

h) the tail type is right- tailed .

i ) the test statistics is :

z = {(x- mu) * sqrt(n) } / sigma ,

where , x= 0.40 , mu = 0.45 , n=16 , sigma= 0.09

so , z = {( 0.40 - 0.45 ) * 4} / 0.09

= - 2.22

j) p value calculated is 0.13142 , which means the test is significant at p < 0.10

k) finally , we say that , we reject null hypothesis and conclude that , the training can increase mean run distance for all dogs .

I HOPE I WAS HELPFUL AND HAVE CLEARED YOUR DOUBTS.

THANKYOU