Question

ot camp. Summary statistics on BMI maasuraments are shown in thetabla to tha right Camplato partsathrough h Mesn Deviation StDa tha diaancs in EMI vaues naad to be nlly distibutad in order far tha interence, part t, to be valid Explsin Choose the cor

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Answer #1

Here, we have given that

n= number of campers =74

\bar x1= mean starting BMI = 34.4

\bar x2=mean ending BMI= 30.7

S1= starting BMI standard deviation = 7.1

S2= ending BMI standard deviation = 6.4

\bar d= Mean difference= 3.7

S.D(diff) =standard deviation of difference = 1.5

Claim: To check whether the mean BMI at the end of camp is less than the mean BMI at the end of camp.

The Hypothesis is

Ho: \mu d= 0

v/s

H1: \mu d < 0

Here, we use paired difference t test because the assumptions of independent sample is invalid

and data is in pre and post format.

Now,

Now, we can find the test statistic

, stat = sd

2 7 , stat = V(74)

=21.22

we get test statistic is 21.22

the test statistics using the formula for paired difference t-test

Degrees of freedom = n-1= 74-1=73

\alpha= level of significance=0.01

Now, we can find the P-value

P-value = 0.0000 ( using EXCEL =TDIST(x,D.F,tail=1))

Decision:

P-value < 0.01 (\alpha)

Conclusion:

That is we fail to reject Ho (Null Hypothesis) because P-value < 0.01 (\alpha)

There is not sufficient evidence that the mean BMI at the end of camp is less than the mean BMI at the end of camp.

9. option A is correct that Yes , in order to make the valid small sample inferences about \mu_d the differences need to be normally distributed.

(B)

Now,

Formula for CI is as follows

In)

First we find the critical value

Degrees of freedom = n-1 = 74-1 = 73

\alpha= level of significance=0.01

We get

Tc= 2.645 ( using t table)

Now,

In)

-2.645 * _15 (74) 1.5 ud 3.7+2615 V(74)

2.24く,ld < 5.16

Interpretation:

This confidence interval shows that we are 99% confident that the difference in the population mean will falls under this interval.

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