Math / Data & Statistics

QuestionFirstborn \begin{tabular}{lllll} \hline 99 & 116 & 103 & 123 & 110 \\ 104 & 94 & 92 & 82 & 85 \\ \hline \end{tabular} \begin{tabular}{lllll} \hline \multicolumn{4}{c}{ Secondborn } \\ \hline 103 & 96 & 120 & 120 & 123 \\ 99 & 110 & 99 & 101 & 108 \\ \hline \end{tabular} Send data to Excel
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Part 1 of 2
Construct a 99.8\% confidence interval for the difference in mean IQ between firstborn and secondborn sons. Let $\mu_{1}$ denote the mean IQ of the firstborn sons. Use tables to find the critical value and round the answers to at least one decimal place.
A 99.8\% confidence interval for the difference in mean IQ between firstborn and secondborn sons is $\square$

Studdy Solution

STEP 1

1. We have two independent samples: Firstborn and Secondborn.
2. The sample sizes for Firstborn and Secondborn are equal.
3. We assume the samples are normally distributed.
4. We will use the t-distribution to find the critical value for the confidence interval.

STEP 2

1. Calculate the sample means and standard deviations for both groups.
2. Determine the standard error of the difference in means.
3. Find the critical value for a 99.8% confidence interval.
4. Calculate the confidence interval for the difference in means.

STEP 3

Calculate the sample mean for the Firstborn group.
Firstborn data: $99, 116, 103, 123, 110, 104, 94, 92, 82, 85$
$\bar{x}_1 = \frac{99 + 116 + 103 + 123 + 110 + 104 + 94 + 92 + 82 + 85}{10}$
$\bar{x}_1 = \frac{1008}{10} = 100.8$

STEP 4

Calculate the sample mean for the Secondborn group.
Secondborn data: $103, 96, 120, 120, 123, 99, 110, 99, 101, 108$
$\bar{x}_2 = \frac{103 + 96 + 120 + 120 + 123 + 99 + 110 + 99 + 101 + 108}{10}$
$\bar{x}_2 = \frac{1079}{10} = 107.9$

STEP 5

Calculate the standard deviation for the Firstborn group.
$s_1 = \sqrt{\frac{\sum (x_i - \bar{x}_1)^2}{n-1}}$
$s_1 = \sqrt{\frac{(99-100.8)^2 + (116-100.8)^2 + \ldots + (85-100.8)^2}{9}}$
$s_1 = \sqrt{\frac{(3.24 + 233.64 + 5.76 + 499.84 + 84.64 + 10.24 + 46.24 + 77.44 + 353.44 + 250.24)}{9}}$
$s_1 = \sqrt{\frac{1564.72}{9}} \approx 13.2$

STEP 6

Calculate the standard deviation for the Secondborn group.
$s_2 = \sqrt{\frac{\sum (x_i - \bar{x}_2)^2}{n-1}}$
$s_2 = \sqrt{\frac{(103-107.9)^2 + (96-107.9)^2 + \ldots + (108-107.9)^2}{9}}$
$s_2 = \sqrt{\frac{(24.01 + 140.49 + 148.84 + 148.84 + 228.01 + 76.41 + 0.01 + 76.41 + 48.41 + 0.01)}{9}}$
$s_2 = \sqrt{\frac{891.44}{9}} \approx 9.9$

STEP 7

Calculate the standard error of the difference in means.
$SE = \sqrt{\frac{s_1^2}{n} + \frac{s_2^2}{n}}$
$SE = \sqrt{\frac{13.2^2}{10} + \frac{9.9^2}{10}}$
$SE = \sqrt{\frac{174.24}{10} + \frac{98.01}{10}}$
$SE = \sqrt{17.424 + 9.801}$
$SE = \sqrt{27.225} \approx 5.2$

STEP 8

Find the critical value for a 99.8% confidence interval with $n_1 + n_2 - 2 = 18$ degrees of freedom.
Using a t-table, the critical value $t^*$ for 99.8% confidence and 18 degrees of freedom is approximately 3.922.

STEP 9

Calculate the confidence interval for the difference in means.
$CI = (\bar{x}_1 - \bar{x}_2) \pm t^* \times SE$
$CI = (100.8 - 107.9) \pm 3.922 \times 5.2$
$CI = -7.1 \pm 20.38$
$CI = (-27.48, 13.28)$
The 99.8% confidence interval for the difference in mean IQ between firstborn and secondborn sons is:
$\boxed{(-27.5, 13.3)}$

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Studdy Solution

STEP 1

1. We have two independent samples: Firstborn and Secondborn.
2. The sample sizes for Firstborn and Secondborn are equal.
3. We assume the samples are normally distributed.
4. We will use the t-distribution to find the critical value for the confidence interval.

STEP 2

1. Calculate the sample means and standard deviations for both groups.
2. Determine the standard error of the difference in means.
3. Find the critical value for a 99.8% confidence interval.
4. Calculate the confidence interval for the difference in means.

STEP 3

Calculate the sample mean for the Firstborn group.
Firstborn data: $99, 116, 103, 123, 110, 104, 94, 92, 82, 85$
$\bar{x}_1 = \frac{99 + 116 + 103 + 123 + 110 + 104 + 94 + 92 + 82 + 85}{10}$
$\bar{x}_1 = \frac{1008}{10} = 100.8$

STEP 4

Calculate the sample mean for the Secondborn group.
Secondborn data: $103, 96, 120, 120, 123, 99, 110, 99, 101, 108$
$\bar{x}_2 = \frac{103 + 96 + 120 + 120 + 123 + 99 + 110 + 99 + 101 + 108}{10}$
$\bar{x}_2 = \frac{1079}{10} = 107.9$

STEP 5

STEP 6

STEP 7

Calculate the standard error of the difference in means.
$SE = \sqrt{\frac{s_1^2}{n} + \frac{s_2^2}{n}}$
$SE = \sqrt{\frac{13.2^2}{10} + \frac{9.9^2}{10}}$
$SE = \sqrt{\frac{174.24}{10} + \frac{98.01}{10}}$
$SE = \sqrt{17.424 + 9.801}$
$SE = \sqrt{27.225} \approx 5.2$

STEP 8

Find the critical value for a 99.8% confidence interval with $n_1 + n_2 - 2 = 18$ degrees of freedom.
Using a t-table, the critical value $t^*$ for 99.8% confidence and 18 degrees of freedom is approximately 3.922.

STEP 9

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Studdy Solution

STEP 1

1. We have two independent samples: Firstborn and Secondborn.2. The sample sizes for Firstborn and Secondborn are equal.3. We assume the samples are normally distributed.4. We will use the t-distribution to find the critical value for the confidence interval.

STEP 2

1. Calculate the sample means and standard deviations for both groups.2. Determine the standard error of the difference in means.3. Find the critical value for a 99.8% confidence interval.4. Calculate the confidence interval for the difference in means.

STEP 3

STEP 4

STEP 5

STEP 6

STEP 7

STEP 8

Find the critical value for a 99.8% confidence interval with n1+n2−2=18 n_1 + n_2 - 2 = 18 n1​+n2​−2=18 degrees of freedom.Using a t-table, the critical value t∗ t^* t∗ for 99.8% confidence and 18 degrees of freedom is approximately 3.922.

STEP 9

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. We have two independent samples: Firstborn and Secondborn.
2. The sample sizes for Firstborn and Secondborn are equal.
3. We assume the samples are normally distributed.
4. We will use the t-distribution to find the critical value for the confidence interval.

1. Calculate the sample means and standard deviations for both groups.
2. Determine the standard error of the difference in means.
3. Find the critical value for a 99.8% confidence interval.
4. Calculate the confidence interval for the difference in means.

Find the critical value for a 99.8% confidence interval with $n_1 + n_2 - 2 = 18$ degrees of freedom.
Using a t-table, the critical value $t^*$ for 99.8% confidence and 18 degrees of freedom is approximately 3.922.