Question

The null and alternate hypotheses are:

H₀ : μ₁ = μ₂
H₁ : μ₁ ≠ μ₂

A random sample of 9 observations from one population revealed a sample mean of 24 and a sample standard deviation of 3.7. A random sample of 6 observations from another population revealed a sample mean of 28 and a sample standard deviation of 4.6.

At the 0.01 significance level, is there a difference between the population means?

a. State the decision rule. (Negative values should be indicated by a minus sign. Round your answers to 3 decimal places.)

b. Compute the pooled estimate of the population variance. (Round your answer to 3 decimal places.)

c. Compute the test statistic. (Negative value should be indicated by a minus sign. Round your answer to 3 decimal places.)

d. State your decision about the null hypothesis.​

e.The  p-value is​

Answer 1

Here we have given that,

Claim: To Check whether the two population mean are equal or not.

n 1 =1^st sample size= 9

= 1^st sample mean =24

S1 = 1^st sample standard deviation = 3.7

n2 = 2^nd sample size= 6

= 2^nd sample mean = 28

S2 = 2^nd sample standard deviation =4.6

The hypothesis is

v/s

To check this two population, mean equal or not 1^st we need to check the population variance is equal or not.

Claim: To check whether the two population variance equal or not.

The hypothesis is

v/s

Test statistics is

  

=0.62

Now,we find the P-value   

=level of significance=0.01

Degrees of freedom 1=n1-1=9-1=8

Degrees of freedom 2= n2-1= 6-1=5

P-value =0.7222 Using EXCEL=FDIST(F-STAT=0.62,D.F1=8,D.F2=5))

Decision:

Here P-value > 0.01

Conclusion:

Then we fail to reject Null hypothesis that the population variances are equal.

Now, We find the test statistics for two sample t test

(A)

Decision Rule:

IF P-value < 0.01 then we reject Ho Null Hypothesis

Otherwise we fail to reject the Null hypothesis.

Test statistics is:

Where Sp =pooled sample variance

(B)

Now we find it

=16.56

Now we get test statistics is

= -1.865

we get Test statistics = -1.865

Now we find the P-value

Degrees of freedom = n1+n2-2 = 9+6-2 =13

This is two tailed test

Now we find the P-value

P-value = 0.1158 (using Excel = TDIST(| t-stat |=1.865, D.F=13, Tail=2))

Decision:

Here P-value > 0.01

That is we Fail to reject Ho Null hypothesis

Conclusion:

That is we say that there is No strong evidence that the two population mean are not equal.