Question

A dairy scientist is testing a new feed additive. She chooses 13 cows at random from a large population. She randomly assigns nold = 8 to the old diet and nnew = 5 to a new diet including the additive. The cows are housed in 13 widely separated pens. After two weeks, she milks each cow and records the milk produced in pounds: Old Diet: 43, 51, 44, 47, 38, 46, 40, 35 New Diet: 47, 75, 85, 100, 58 Let μnew and μold be the population mean milk productions for the new and old diets, respectively. She wishes to test H0 : μnew − μold = 0 against HA : μnew − μold ̸= 0 using α = 0.05.(a) Graph the data as you see fit. Why did you choose the graph(s) that you did and what does it (do they) tell you?(b) Perform the hypothesis test assuming equal population variance. Compute the p-value and make a reject or not reject decision. State your conclusion in the context of the problem. (c) Repeat the previous part, but without the equal variance assumption. (d) Compare the results from part b and c. Which test do you trust more and why? I have answers for (a) and (b) with p-value 0.035 which rejects null hypothesis. Can you help with (c) and (d)? Thank you!

Answer 1

Here we have given that,

Claim: To Check whether the two population mean of old diet of cow and new diet of cow are equal or not.

Old Diet (X1)	New Diet (X2)
43	47
51	75
44	85
47	100
38	58
46
40
35

n 1 =number of cows taking old diet= 8

n2 = number of cows taking new diet=5

= sample mean of old Diet  =43

= sample mean of new diet = 73

S1 =sample standard deviation of old diet =5.18

S2 = sample standard deviation of new diet=21.08

The hypothesis is

v/s

(A)

Here both graphs shows the all the data points are within the reference line so, bothe data follwows the normal distribution

(B)

Now,

We assume that the two populaiton variances are equal.

Test statistics is:

Where Sp =pooled sample variance

Now we find it

=178.66

Now we get test statistics is

= -3.94

we get

Test statistic is -3.94

Now we find the P-value

Degrees of freedom = n1+n2-2 =8+5-2 =11

=level of significance=0.05

P-value= 0.0024 Using Excel =TDIST( | T-STATISTICS |=3.94, d.f=11, TAIL =2)

Decision:

Here P-value < 0.05

That is we reject Ho Null hypothesis

That is we say that there is strong evidence that the two population mean of old diet of cow and new diet of cow are equal.

(C)

Now, we want to repeat previous part Assuming the two population variances are Not equal.

we perform this using excel softaware

Data << Data anlaysis <<< 2 sample t using unequal variances<<

select variables and output range

we get the following output

t-Test: Two-Sample Assuming Unequal Variances
	Old Diet (X1)	New Diet (X2)
Mean	43.00	73.00
Variance	26.86	444.50
Observations	8	5
Hypothesized Mean Difference	0
df	4
t Stat	-3.123
P(T<=t) one-tail	0.018
t Critical one-tail	2.132
P(T<=t) two-tail	0.035
t Critical two-tail	2.776

Here we get

Test statistis :

T-statistics= -3.123

and

P-value = 0.035

Decision

Here P-value < 0.05

That is we reject Ho Null hypothesis

Conclusion:

That is we say that there is strong evidence that the two population mean of old diet of cow and new diet of cow are equal.