can you show your work how to do these? also, I don\'t understand how to get the

Question

can you show your work how to do these? also, I don't understand how to get the p value. thanks.

You may need to use the appropnate table in Appendix is to answer Uus question 6 O 010 points Anowers JKEStat11 10 E 072 Twenty laboratory mice were randomly divided into two groups of 10. Each group was fed according to a prescribed diet. At the end of 3 weeks, the weight gained by each animal was recorded. Do the data in the following table justify the conclusion that the mean weight gained on diet B was greater than the mean weight gained on diet A, at the a 0.oS level of significance? Assume normality (Use Diet B - Diet A.) Diet A 7 9 13 96 146 7 12 12 Diet B 8 10 21 21 11 8 17 611 14 a) Find t. (Give your anawer correct to two decimal places (ii) Find the p-value. (Give your answer correct to four decimal places.) State the D Reject the null hypothesis, there is significant evidence that diet B had a greater weight gain, @) Reject the null hypothesis, there is not significant evidence that diet B had a greater weight gain. Fail to reject the null hypothesis, there is significant evidence that diet B had a greater weight gain. Fail to reject the null hypothesis, there is not significant evidence that diet B had a greater weight gain. You may need to use the appropriate table in Appendix B to answer this question Type here to search

Explanation / Answer

Solution

Let xi = weight gained by the ith mouse fed with diet A

Let yi = weight gained by the ith mouse fed with diet B

n is the sample size common to both A and B.

Then, Xbar = sample mean of X and Ybar = sample mean of Y

s12 = sample variance of X and s22 = sample variance of Y.

Let s2 = (s12 + s22)/2

Part (a)(i)

Test statistic: t = (Xbar - Ybar)/s(2/n)

From the given data the following are obtained using Excel Functions

Xbar

9.5

Ybar

12.7

s1^2

149.5556

s2^2

28.9

s^2

89.22778

s

9.446046

t

- 0.7575 ANSWER

Part (a)(ii)

The above t has t-distribution with degrees of freedom = 2(n - 1) = 18.

So, p-value = P(t18 < - 0.7575) = 0.2293 [obtained using Excel Function] ANSWER

Method to find p-value using Excel Function for t-Distribution

In Excel Worksheet, click ‘fX’. Screen displays ‘Insert Function’ In the window ‘Search for a function’ type ‘TDIST’ and click ‘Go’. Under ‘Select a function’, TDIST appears. Click ‘Ok’

Screen displays three windows. Against ‘X’ window, type the mod of calculated value (i.e., without - sign) of the test statistic. Against the ‘Deg_freedom’ type the degrees of freedom and in the other window ‘Tails’ type 1

Required probability, i.e., the p-value, is displayed below the windows against ‘=’ sign.

The above method is illustrated below

tcal

| tcal |

DF

Prob = p-value

-0.7575

0.7575

18

0.229276453

Part (b)

Since p-value is very high, even greater than 10% (i.e., 0.1), we accept the null hypothesis that there is no difference between the mean of gain in weights of the two diets. This => that we fail to reject null hypothesis that diet B shows a greater weight gain

The last option ANSWER

Xbar

9.5

Ybar

12.7

s1^2

149.5556

s2^2

28.9

s^2

89.22778

s

9.446046

t

- 0.7575 ANSWER

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