Question

4. Train for Scooby. Organizations such as PETA struggle to tackle the deficiency of competent trainers by employing uncertified trainers in dog hounds with critical needs often in remote areas. A 2019 study compared - r-old dogs trained by certified trainers with others trained by under-certified trainers in the same remote areas. Obedience of under-certified trainers averaged 36.52 points with standard deviation 9.31. The obedience of 2-yr-old logs trained by certified trainers had mean 32.48 points with standard deviation 9.43 points on the same test. There vere 44 2-yr-old dogs in each group. Is there evidence of lower scores with under-certified trainers? a) Define the population parameter in this context. Which method is this? b) Are these quantitative or categorical data? coteqorical c) Do these data satisfy the necessary conditions? Explain each condition. Show me the numbers. (independence can be skipped) d) Make a confidence interval and write a sentence interpreting it. You are to write the formula from the cheat sheet, and note each variable in that formula and show work to solve for that variable, including the SE. Then plug in those values to the equation and solve for the CI. Start from left to right as you read the formula so that there is order/consistency to how you do each problem. d) What would you conclude about the two types of trainers?

Answer 1

(a)

In this context, the population parameter of interest is the difference in the scores of two types of trainers. This is hypothesis testing method.

(b)

These are quantitative data.

(c)

The first condition is the condition of normality. It is given that the data is distributed as normal distribution in both the cases. Hence this condition is satisfied.

The second condition is the condition of heteroscedasticity. This is also satisfied by the data as the respective SDs are given.

(d)

Let the mean population score of certified trainers be $\mu_1$ and that os uncertified trainers be $\mu_2$

Now we make a table along with the symbols.

	Sample 1	Sample 2
Sample size	n1=44	n2=44
Sample mean	C1 = 36.52	T2 = 32.48
Sample SD	s1=9.31	s2=9.43

Now the 95% confidence interval is given by

(21 - 22 S.E.V - + - V ni n2 t 0.025,71 +92-2)

where S.E. is given by

$S.E.=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}$

Here,

$S.E.=\sqrt{\frac{(44-1)9.31^2+(44-1)9.43^2}{44+44-2}}$

$\Rightarrow S.E.=\sqrt{\frac{7550.843}{86}}=9.3701921$

Hence the CI is given by

$((36.52-32.48)\pm 9.3701921 \sqrt{\frac{1}{44}+\frac{1}{44}}\;\;t_{\;0.025,86})$

$i.e. (4.04\pm (9.3701921*0.2132007164*1.99))$

$i.e. (4.04\pm 3.97548602)$

(d)

The difference between the scores of the two trainers would lie between 0.06451398046 and 8.01548062

(e)

The S.E. is obtained as 9.3701921

1.

The critical value for alpha=0.05 is obtained as

$t_{0.025,86}\simeq 1.99$

2.

The test statistic is given by

$t=\frac{\bar{x}_1-\bar{x}_2}{S.E.\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

From the data, it is given by

$t=\frac{36.52-32.48}{9.3701921\sqrt{\frac{1}{44}+\frac{1}{44}}}$

$\Rightarrow t=\frac{4.04}{1.997731668}=2.022293617$

4.

The p-value is obtained from the table as 0.023126

5.

As the p-value is less than 0.05, the null is rejected at 5% level of significance and hence we can conclude that there is evidence of low scores with under-certified trainers.

Hopefully this will help you. In case of any query, do comment. If you are satisfied with the answer, give it a like. Thanks.