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N76 [4]
3 years ago
6

A dairy scientist is testing a new feed additive. She chooses 13 cows at random from a large population of cows. She randomly as

signs nold = 8 to get the old diet, and nnew = 5 to get the new diet including the additive. The cows are housed in 13 separated pens and each gets separate feed, with or without additive as appropriate. After two weeks, she picks a day and milks each cow using standard procedures and records the milk produced in pounds. The data are below:
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 vs. HA : µnew ? µold 6= 0, using ? = 0.05.

(a) Are the two populations paired or independent? Explain your answer.

(b) 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?

(c) Choose a test appropriate for the hypotheses above, and justify your choice based on your answers to parts (a) and (b). Then perform the test by computing a p-value, and making a reject or not reject decision. Do not use R for this, and show your work. Finally, state your conclusion in the context of the problem.

Mathematics
1 answer:
mina [271]3 years ago
3 0

Answer:

Step-by-step explanation:

Given that a dairy scientist is testing a new feed additive. She chooses 13 cows at random from a large population of cows. She randomly assigns nold = 8 to get the old diet, and nnew = 5 to get the new diet including the additive.

From the data given we get the following

           N       Mean      StDev     SE Mean

Sample   1    8           43       5.1824 1.832

Sample   2   5           73     21.0832 9.429

df = 11

Std dev for difference = 13.3689

a) Yes the two are independent.  The two sets of cows randomly chosen are definitely independent. Paired means equal number should be there and homogeneous conditions should be maintained.

b) Enclosed

c) Comparison of two means is the test recommended here.  Because independent samples are used.\

d) Test statistic= -3.1233

(because of unequal variances we use that method)

95% confidence interval =  ( -56.6676 , -3.3324 )

p value <0.05 our alpha

So reject null hypothesis.

The two means are statistically significantly different.

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Robert is inspecting a shipment of 22 inch pipes. The length of the pipes may vary by 1%. What is the range of allowable lengths
Slav-nsk [51]

Answer:

The allowable lengths of pipe is from 21.78 inch to 22.22 inch.

Step-by-step explanation:

The length of the each pipe in the shipment =  22 inch

The variation in the length of pipes  = 1%

⇒The allowable length of pipe in the inspection  = 22 inch ± 1%

Now, 1% of 22 = \frac{1}{100}  \times 22 = 0.22

So,  22 inch + 1% = 22 inch + 0.22 =  22.22 inch

and 22 inch -  1% = 22 inch - 0.22 =  21.78 inch

⇒ 22 inch ± 1%  =  22.22 inch and 21.78 inch

Hence, the allowable lengths of pipe is from 21.78 inch to 22.22 inch.

3 0
2 years ago
consider the following scenario: Assume daily protein intake values in a population of athletes are known to have a standard dev
Kipish [7]

Answer:

The 95% confidence interval for the population mean daily protein intake is between 69.97g and 84.03g.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.96*\frac{58.6}{\sqrt{267}} = 7.03

The lower end of the interval is the sample mean subtracted by M. So it is 77 - 7.03 = 69.97g.

The upper end of the interval is the sample mean added to M. So it is 77 + 7.03 = 84.03g.

The 95% confidence interval for the population mean daily protein intake is between 69.97g and 84.03g.

5 0
3 years ago
Nicholas wrote the steps below to simplify the fraction 20/30. Find his error and correct it. 20/30= 20÷5=4 30÷6=5
aksik [14]

Answer:2/3

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
400.0 ml of a gas are under pressure of 80.0 kpa what would the volume of the gas be at a pressure of 100.0 kpa
Ber [7]

Answer:

The final volume is 320 mL at pressure of 100 kPa.

Step-by-step explanation:

Boyle's law gives the relation between volume and pressure of a gas. It states that at constant temperature, volume is inversely proportional to its pressure such that,

V\propto \dfrac{1}{P}\\\\P_1V_1=P_2V_2

Let

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We need to find V_2. Using above equation, we get :

V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{400\times 80\times 10^3}{100\times 10^3}\\\\V_2=320\ mL

So, the final volume is 320 mL at pressure of 100 kPa.

6 0
3 years ago
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