Question

The National Prenatal Statistics Unit of the Sydney Children’s Hospital in reports that the mean birth weight of all babies born in Australia in 2011 was 7.48 pounds.

A Florida hospital reports that the average weight of 118 babies born there last year was 7.52 pounds, with a standard deviation of 1.31 pounds.

If we believe that Florida babies fairly represent American newborns, is there any evidence that U.S. babies and Australian babies do not weigh the same amount at birth?

The Hypotheses for the test are

H_O : μ = 7.48

H_A : μ ≠ 7.48

1. Find the P-value for the test. (Due to the different methods, round to 2 decimal places and select the best possible answer)

P-value:                            [ Select ]                       ["0.04", "0.21", "0.03", "0.74"]      

2. Using an α = 0.10 significance level and your P-value , we would conclude:

Decision:

Conclusion:

3. What would be a Type I Error in the context of this problem?   

4. Find a 90% confidence interval for the true mean weight of American newborns

Answer 1

Solution-1:

t=xbar-mu/s/sqrt(n)

=(7.52 -7.48)/(1.31/sqrt(118))

t=0.331688

df=n-1=118-1=117

two tail p value in excel

=T.DIST.2T(0.331688,117)

=0.74071827

p=0.74

2. Using an α = 0.10 significance level and your P-value , we would conclude:

Decision:p=0.74>0.10

Conclusion:Fail to reject Ho

3. What would be a Type I Error in the context of this problem?   

Type 1 error=alpha=0.10

Solution-4:

df=n-1=118-1=117

alpha=0.10

alpha/2=0.10/2=0.05

t crit in excel

==T.INV(0.05,117)

=1.657981659

90% confidence interval for mean

xbar-tc*s/sqrt(n),xbar+tc*s/sqrt(n)

7.52-1.657981659*1.31 /sqrt(118),7.52+1.657981659*1.31 /sqrt(118)

7.320055, 7.719945

we are 90% confident that the true mean weight of  American newborns lies in between 7.320055 and 7.719945