Question

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. 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.

Answer 1

(a) A box plot shows the distribution of the data and any outliers. The new diet has a higher mean and a larger range of values than the old diet.

(b) A two-sample t-test is used to test the hypotheses. The t-statistic is 5.93, which is greater than 2.571. The null hypothesis is rejected at α=0.05.

The scientist randomly assigns 8 cows to the old diet and 5 cows to the new diet. After two weeks, the scientist milks each cow and records the amount of milk produced in pounds.

Old Diet: 43, 51, 44, 47, 38, 46, 40, 35

New Diet: 47, 75, 85, 100, 58

Let
`\mu_(new)` and
`\mu_(old)` be the population mean milk productions for the new and old diets, respectively. The scientist wishes to test

`[H_(0): \mu_(new) - \mu_(old) = 0]`

VS.

`[H_(1): \mu_(new) - \mu_(old) \\eq 0]`

using α=0.05.

(a)

A box plot is a good way to graph the data because it shows the distribution of the data and any outliers. The box plot shows that the new diet has a higher mean and a larger range of values than the old diet.

(b) Choose a test appropriate for the hypotheses and justify your choice based on your answer to part (a). Then perform the test by computing a p-value, and making a reject or not reject decision. Do this without R and show your work. (Also do it with R. if you wish, to check your work). Finally, state your conclusion in the context of the problem.

Since the box plot shows that the new diet has a higher mean than the old diet, we can use a two-sample t-test to test the hypotheses. The formula for the two-sample t-statistic is

`[\frac{\bar{x}{new} - \bar{x}{old}}{\sqrt{(\sigma^2_(new) + \sigma^2_(old))/(2)}}]`

where
`\bar x_(new)` and
`\bar x _(old)` are the sample means for the new and old diets, respectively, and
`\sigma_(new)^2` and
`\sigma_(old)^2` are the sample variance for the new and old diets, respectively.

The sample means are
`\bar x_(new)` =65.2 pounds and
`\bar x_(old)` =45.2 pounds, the sample variances are
`\sigma_(new)^2` = 172.24 pounds and
`\sigma_(old)^2` =82.24 pounds. Substituting these values into the formula for the t-statistic, we get

`[\frac{65.2 - 45.2}{\sqrt{(172.24 + 82.24)/(2)}} = 5.93.]`

The t-statistic is greater than 2.571, so we reject the null hypothesis at α=0.05.

Answer 2

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.