Question

8.        Random sampling from two normal populations produced the following results:

x₁ = 412       s₁ = 128      n₁ = 150

x₂ = 405     s₂ = 54       n₂ = 150

(a.) Can we infer at the 5% significant level that m₁

is greater than m₂ ?

(b.) Repeat part (a) decreasing the standard deviations to: s₁ = 31 and s₂ = 16

 

 

9.            In random samples of 25 from eah of two normal populations we found the following statistics:

x₁ = 524      s₁ = 129

x₂ = 469     s₂ = 141

(a.) Estimate the difference between the two population means with 95% confidence (b.) Repeat part (a) increasing the standard deviations

to: s₁ = 255 and s₂ = 260

Answer 1

8 (a) Null hypothesis: $m{\scriptscriptstyle 1}≤m{\scriptscriptstyle 2}$

Alternative hypothesis: $m{\scriptscriptstyle 1}>m{\scriptscriptstyle 2}$

Since both sample sizes are greater than 30, we can use Z-score

According to the form of the null hypoyhesis, we should use right-tailed test

$P(Z>Z{\scriptscriptstyle cr})=0.05\implies Z{\scriptscriptstyle cr}=1.645$

The test value is:

$Z{\scriptscriptstyle test}={\frac {(x{\scriptscriptstyle 1}-x{\scriptscriptstyle 2})} {\sqrt{({\frac {s^2{\scriptscriptstyle 1}} {n{\scriptscriptstyle 1}}} + {\frac {s^2{\scriptscriptstyle 2}} {n{\scriptscriptstyle 2}})}}}}={\frac {412-405} {\sqrt{{\frac {128^2} {150}}+{\frac {54^2} {150}}}}}={\frac 7 {11.3}}=0.62$

There is not enoughg evidencve to reject null hypothesis on th 0.05 significance level, so m₁

is not greater than m₂

(b) Now s₁ = 31 and s₂ = 16, which means our test statistic would be

$Z{\scriptscriptstyle test}={\frac {(x{\scriptscriptstyle 1}-x{\scriptscriptstyle 2})} {\sqrt{({\frac {s^2{\scriptscriptstyle 1}} {n{\scriptscriptstyle 1}}} + {\frac {s^2{\scriptscriptstyle 2}} {n{\scriptscriptstyle 2}})}}}}={\frac {412-405} {\sqrt{{\frac {31^2} {150}}+{\frac {16^2} {150}}}}}={\frac 7 {2.85}}=2.456$

There is enoughg evidencve to reject null hypothesis on th 0.05 significance level, so m₁

is greater than m₂

9 (a) The difference between two sample means with 95% confidence is

$a \in ($$(x{\scriptscriptstyle 1} - x{\scriptscriptstyle 2})-Cr{\scriptscriptstyle 0.95}*{\sqrt{({\frac {s^2{\scriptscriptstyle 1}} {n{\scriptscriptstyle 1}}} + {\frac {s^2{\scriptscriptstyle 2}} {n{\scriptscriptstyle 2}})}}};(x{\scriptscriptstyle 1} - x{\scriptscriptstyle 2})+Cr{\scriptscriptstyle 0.95}*{\sqrt{({\frac {s^2{\scriptscriptstyle 1}} {n{\scriptscriptstyle 1}}} + {\frac {s^2{\scriptscriptstyle 2}} {n{\scriptscriptstyle 2}})}}})$

Since we have small sample sizes, then it is appropriate to use t-statistic(two-tailed) with n - 1 degrees of freedom

$P(T(24)>Cr{\scriptscriptstyle 0.95})=0.025\implies Cr = 2.063$

$(x{\scriptscriptstyle 1} - x{\scriptscriptstyle 2})*Cr{\scriptscriptstyle 0.95}*{\sqrt{({\frac {s^2{\scriptscriptstyle 1}} {n{\scriptscriptstyle 1}}} + {\frac {s^2{\scriptscriptstyle 2}} {n{\scriptscriptstyle 2}})}}})=(55-2.063*38.2;55+2.063*38.2)=(-23.8;133.8)$

(-23.8;133.8) is the 95% confidence interval

(b) for s₁ = 255 and s₂ = 260 we have

$(x{\scriptscriptstyle 1} - x{\scriptscriptstyle 2})*Cr{\scriptscriptstyle 0.95}*{\sqrt{({\frac {s^2{\scriptscriptstyle 1}} {n{\scriptscriptstyle 1}}} + {\frac {s^2{\scriptscriptstyle 2}} {n{\scriptscriptstyle 2}})}}})=(55-2.063*72.84;55+2.063*72.84)=(-95.27;205.29)$

(-95.27;205.29) is tha 95% confidence interval