Question

levels of serotonin were measured in a sample of 25 dogs in their puppy years. The sample mean and the sample standard deviation were 73 and 16, respectively. Can it be concluded from the data that the population mean level is greater than 70? Use α = 0.05 level of significance.

a what is the value for the best point estimate for the circulating levels of estrogen in the population?
b What assumption(s) must be made?
c State the hypotheses and identify the claim.
d Find the critical value(s) and the critical region(s).
e Compute the test value.
f Make the decision to reject or not reject the null hypothesis.
g Summarize the results.
h Compute the p-value for the test statistic.

Answer 1

Solution:

a)

Point estimate is,

sample mean is = 73

b) We assume that ,

levels of serotonin follows approximately normal distribution.

Population standard deviation is unknown and sample size is <30.

c)

The claim: The population mean level is greater than 70

Null and Alternative Hypotheses:

d)

The number of degrees of freedom are df = n-1 = 25-1=24

Rejection Region:
Given significance level = α = 0.05 and df = 24

So Critical Value for the test can be found using excel formula, =TINV(2*0.05,24)

tc = 1.711

The rejection region for this right-tailed test is R=t:t>1.711

e)

Test statistic:

Formula:

f)

Decision:

Since it is observed that |t| = 0.938 < tc = 1.711
It is then concluded that the Null Hypothesis is not rejected.

g)

Conclusion:Therefore, there is not enough evidence to claim that the population mean level is greater than 70.

h)

P-value approach:

Using excel formula, =TDIST(0.938,24,1)

P-value = 0.1789

So, P-value >0.05, do not reject H0

Done