Question

Toilet Training You are a great friend and are taking your friend's child to the playground to play with two of his friends: Charles and Elizabeth. You overhear the dads of Charles and Elizabeth talking about a recent study regarding toilet training of children. The study found that the mean age for girls to stay dry during the day (successful completion of toilet training) is 32.5 months, and the mean age for boys is 35.0 months. These two groups had reported standard deviations for the age when a child is successfully toilet trained of 6.7 months for girls and 10.1 months for boys based on a sample of 126 girls and a second sample of 141 boys. Charles's dad and Elizabeth's dad are getting into a disagreement about how to interpret these results. Based on your knowledge from Stats 250, can you help settle their disagreement by helping to answer the following questions about the difference between the population mean age for girls to be successfully toilet trained and the population mean age for boys to be successfully toilet trained? Question 5 Subquestions 5.
a TBD points Based on our given information, should you use the unpooled (Welch's) or the pooled approach to calculate the confidence interval? Make sure to include numerical support for your answer. No answer entered. Click above to enter an answer. 5.

b TBD points The reported 95% confidence interval is (0.3927 months. 4.6073 months). Based on this confidence interval, which group was chosen to be group 1? How do you know? What is the probability that our parameter of interest, the true difference in population means of ages of successful toilet training between boys and girls, is included in the interval? No answer entered. Click above to enter an answer. 5.

C TBD points Charles's dad is upset that this result guarantees that his son will be at a disadvantage, since he will be toilet trained later than Elizabeth. Elizabeth's dad starts to correct him, stating that this only means that there is only a 95% probability that Elizabeth will be toilet trained before Charles. Are either of these statements correct? Explain why you made your choice.

Answer 1

Answer:

Step-by-step explanation:

Hello!

According to a study regarding the average age of female and male kids to complete toilet training is:

Females:

Average age 32.5 months

The standard deviation of 6.7 months

n= 126

Males:

Average age 35.0 months

The standard deviation of 10.1 months

n= 141

The parameter of study is the difference between the age of females are successfully toilet trained and the average age that males are successfully toilet trained. μf - μm (f= female and m= male)

a.

Assuming that both variables have a normal distribution to choose whether you'll use an unpooled or pooled-t to calculate the confidence interval you have to conduct an F-test for variance homogeneity.

If the variances are equal, then you can usee the pooled-t, but if the variances are different, you have to uses Wlche's approach:

H₀: δ²f = δ²m

H₁: δ²f ≠ δ²m

Since both items b. and c. ask for a 95% CI I'll use the complementary significance level for this test:

α: 0.05

$F= \frac{S^2_f}{S^2_m} * \frac{xSigma^2_f}{Sigma^2_m} ~~~F_{(n_f-1); (n_m-1)}$

$F= \frac{(6.7^2)}{(10.1)^2} *1= 0.44$

Critical values:

$F_{125;140;0.025}= 0.71\\F_{125;140;0.975}= 1.41$

The calculated F value is less than the lower critical value, 0.77, so the decision is to reject the null hypothesis. In other words, there is no significant evidence to conclude the population variances of the age kids are toilet trained to be equal. You should use Welch's approach to construct the Confidence Intervals.

$Df_w= \frac{(\frac{S^2_f}{n_f} + \frac{S^2_m}{n_m} )^2}{\frac{(\frac{S^2_f}{n_f} )^2}{n_f-1} +\frac{(\frac{S^2_m}{n_m} )^2}{n_m-1} }$

$Df_w= \frac{(\frac{6.7^2}{126} + \frac{10.1^2_m}{141} )^2}{\frac{(\frac{126^2}{126} )^2}{126-1} +\frac{(\frac{10.1^2}{141} )^2}{141-1} } = 254.32$

b.

The given interval is:

[0.3627; 4.6073]

Using Welch's approach, the formula for the CI is:

(X[bar]f- X[bar]m) ± $t_{Df_w;1-\alpha /2}$ * $\sqrt{\frac{S^2_f}{n_f} +\frac{S^2_m}{n_m} }$

or

(X[bar]m- X[bar]f) ± $t_{Df_w;1-\alpha /2}$ * $\sqrt{\frac{S^2_f}{n_f} +\frac{S^2_m}{n_m} }$

As you can see either way you calculate the interval, it is centered in the difference between the two sample means, so you can clear the value of that difference by:

(Upper bond - Lower bond)/2= (4.6073-0.3627)/2= 2.1223

The average age for females is 32.5 months and for males, it is 35 months.

Since the difference between the sample means is positive, we can say that the boys were considered "group 1" and the girls were considered "group 2"

You have a95% confidence that the parameter of interest is included in the given confidence interval.

c.

None of the statements is correct, the interval gives you information about the difference between the average age the kids are toilet trained, that is between the expected ages for the entire population of male and female babies.

This represents a guideline but is not necessarily true to all individuals of the population since some male babies can be toilet trained before that is expected as some female babies can be toiled trained after the average value.

I hope it helps!