Statistic question 6 Please dont do in Rminilab and excel I

Statistic question 6

Please don\'t do in R,minilab, and excel, I need all steps.

Some soap manufactuiers sell special \"antibacterial\" soaps. However, one might expect ordinary soap to also kill bacteria. To investigate this, a researcher prepared a solution from ordinary, nonantibiotic soap and a control solution of sterile water. The two solutions were placed onto petri dishes and E. coli bacteria were added. The dishes were incubated for 24 hours and the number of bacteria colonies on each dish were counted. The data are given in the following table. Construct a 90% confidence interval for the above data. Be sure to interpret this confidence interval in the context of this setting. A test should be performed to determine whether soap more effective than the control. Conduct the hypothesis test at the 5% level. State the null and alternative hypotheses in words and symbols. Compute the test statistic Compute the P-value State the conclusion of the test in the context of this setting.

Solution

I)
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=41.8
Standard deviation( sd1 )=15.6
Sample Size(n1)=8
Mean(x2)=32.4
Standard deviation( sd2 )=22.8
Sample Size(n1)=7
CI = [ ( 41.8-32.4) ±t a/2 * Sqrt( 243.36/8+519.84/7)]
= [ (9.4) ± t a/2 * Sqrt( 104.68) ]
= [ (9.4) ± 1.943 * Sqrt( 104.68) ]
= [-10.48 , 29.28]
II)

Set Up Hypothesis
Null, soap less effecctive than control Ho: u1 < u2
Alternate, soap more effecctive than control H1: u1 > u2
Test Statistic
X(Mean)=41.8
Standard Deviation(s.d1)=15.6 ; Number(n1)=8
Y(Mean)=32.4
Standard Deviation(s.d2)=22.8; Number(n2)=7
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =41.8-32.4/Sqrt((243.36/8)+(519.84/7))
to =0.92
| to | =0.92
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 6 d.f is 1.943
We got |to| = 0.91873 & | t | = 1.943
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value:Right Tail -Ha : ( P > 0.9187 ) = 0.19683
Hence Value of P0.05 < 0.19683,Here We Do not Reject Ho

We conclude that soap more effecctive than control