Question

Researchers examined the bill color (in hue degree) of male and female zebra finches. Here are the summary statistics from the study. How different are male and female zebra finches in bill color? The margin of error for a 95% confidence interval for the difference in mean bill color μF – μM (in hue degree) is ____. Round your answer to the hundreths decimal place.

Answer 1

Answer:

The answer to this question can be defined as follows:

Step-by-step explanation:

In the question some iformation is missing, which can be defined as follows:

Given:

The value of males:

$n_1= 59\\s_1= 1.46\\\bar y_1=2.91$

The value of Females:

$n_2=60\\s_2= 2.48\\\bar y_2=7.42$

Calculating Degrees of Freedom:

$\bold{ df=\frac{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}{ \frac{1}{n_1-1} (\frac{s_1^2}{n_1})^2+\frac{1}{n_2-1} (\frac{s_2^2}{n_2})^2}}$

    $=\frac{(\frac{1.46^2}{59}+\frac{2,48^2}{60})^2}{ \frac{1}{59-1} (\frac{1.46^2}{59})^2+\frac{1}{60-1} (\frac{2.48^2}{60})^2}$

    $=9.58 or 95$

$\alpha= 1-0.95$

   $= 0.05$

for 95% confidence interval are:

$\frac{\alpha}{2} = \frac{0.05}{2} = 0.025$  

From its t-distribution table , the value of t has an region of ($\frac{\alpha}{2} \ \ 0.025$) for (df=95) in its upper tail.

shall be given by = t {0.025}=1.985

Calculating the margin of Error:

$M_OE =t_{0.025} \times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

         $=1.985\times \sqrt{\frac{1.46^2}{59}+\frac{2.48^2}{60}}\\\\=0.739$

The difference in mean bill color "$\mu_2 -\mu_1$" with 95% confidence intervals:

$\to (7.42 - 2.91) \pm 0.739\\\\\to 4.51 \pm 0.739$

Lower limit=3.771

Upper limit = 5.249