A large-scale study in the 1980s revealed that the mean of theadult Bald Eagle wingspan was about 2.06 meters. Researchers want to evaluate this knowledge using a new sample of the adult Bald Eagles in a hypothesis test at the 5% significance level.

Let be the population mean of the Bald Eagle wingspan measured in meters. The hypotheses are

H0:=2.06

HA: >2.06

The researchers choose a one-sided test because they suspect that the wingspan is getting larger slowly over the generations.

After measuring the wingspan of 64 adults Bald Eagles, the following statistics were computed:

The sample mean is 2.05 m.The sample variance is 0.0049 (s^2).

Please answer the following questions. You may assume that the wingspan of the adult Bald Eagle follows a Normal distribution.

(a)_Compute the value of the test statistic based on the sample of 64 adult Bald Eagles. Please show your work.

t= (x -1)/(sn)

= (2.05-1)/(0.07*8)

=1.05/0.056

18.75

(b)_Assuming that=2.06 (which means that the null hypothesis is true), the test statistic follows a t distribution. What are the degrees of freedom of the t distribution? There is no need to explain.

df=64-1=63

(c)_Is the p-value greater or less than 0.5 (one-half)? Please explain. You are not allowed to compute the value of the p-value.

(d)_Based on Part (c), do you reject the null hypothesis? Please explain.

(e)_Based on this sample of 64 adults Bald Eagles, construct a 90% (two-sided) confidence interval for. Please show your work.

The following information may be useful to you:

Let T be the t distribution with 63 degrees of freedom, P(T < 1.67) = 0.95, P(T< 1.30) = 0.90