Consider the following competing hypotheses and accompanying sample data drawn i

Question

Consider the following competing hypotheses and accompanying sample data drawn independently from normally distributed populations. (Note: the automated question following this one will ask you confidence interval questions for this same data, so jot down your work.)

Calculate the value of the test statistic. (Negative values should be indicated by a minus sign. Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.)

Yes, since the value of the test statistic is less than the critical value of -1.645.

Now provide confidence interval information from the previous question. Specifically:

a.    What is the value of the point estimate of the difference between the two population means?

b.    What is the margin of error at 90% confidence?

        (± what value; please provide to 4 decimals; e.g. "3.1234")

c.    With that margin of error, what is the low number in the confidence interval?

d.    With that margin of error, what is the high number in the confidence interval?

Consider the following competing hypotheses and accompanying sample data drawn independently from normally distributed populations. (Note: the automated question following this one will ask you confidence interval questions for this same data, so jot down your work.)

H0: ?1 ? ?2 = 0 HA: ?1 ? ?2 ? 0

Explanation / Answer

Hypothesis testing

A1)

Sp^2
{(n1-1)*s1^2 + (n2-1)*s2^2}/{n1+n2-2}
=((16-1)*1.68^2+(16-1)*11^2)/(16+16-2)
61.9112

T=
(Xbar1-Xbar2)/sqrt(Sp^2*(1/n1+1/n2))
(79-73)/SQRT(0.1764*(1/16+1/16))
2.16

A2)

P-value
2*(1-P(T<|t|)
2*(1-P(T<abs(2.1568))
T.DIST.2T(abs(2.1568),30)
0.0394

A3)

Yes, since the p-value is less than ?.

B)

T(a/2, df)
=t(0.1/2,30)
=t.INV.2T(0.1,30) =
1.6973

Yes, since the value of the test statistic is less than the critical value of-1.96.

Confidence interval

A) point estimate = (X1bar-X2bar) = 6

b) ME=t(a/2,df)*sqrt(Sp^2*(1/n1+1/n2))
=1.697*SQRT(61.9112*(1/16+1/16))
=4.7209

c) lower =6-4.7209 = 1.2791

d) upper =6+4.7209 = 10.7209

Sample 1 Sample 2 n= 16.000 16.000 mean= 79.000 73.000 s= 1.680 11.000 s^2/n 0.176 7.563

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