I just need help with d,e,fi attached the work for a,b,c A researcher conducts a study to determine whether spending time outdoors improves happiness. She randomly assigns 42 participants into two groups. One group serves as the control group, while the treatment group is scheduled for outdoor activities 4 days per week. The researcher measures the happiness of each participant and the data reveal the following at the conclusion of her study:- Treatment Group: $ n = 21, M_1 = 8, SS = 9 $- Control Group: $ n = 21, M_2 = 6, SS = 7 $What type of design is this study (single-sample, independent measures, repeated measures)?(a) State the null and alternative hypotheses(b) Using an alpha level of 0.05 ($ \alpha = 0.05 $), identify the critical t-value for this hypothesis test.(c) Compute the standard error for this statistical analysis AND describe what this number tells you specifically for this type of design(d) Compute the t-statistic for this hypothesis test and describe what the t-score tells you(e) Determine whether the outdoor exposure program increased happiness, explaining your decision(f) Compute Cohen’s d and describe what this number tells you in this specific study **Hypotheses:**- $ H_0: M_1 = M_2 $- $ H_1: M_1 \neq M_2 $**1. Degree of Freedom:**- Calculated as $ 21 + 21 - 2 = 40 $.**2. Critical Values:**- Alpha level ($ \alpha $) = 0.05 for a two-tailed test.- Two tails critical value = $ \pm 2.021 $.**3. Pooled Standard Deviation:**- Formula: \[ \sqrt{\frac{(21 - 1)9^2 + (21 - 1)7^2}{21 + 21 - 2}} \]- Calculated as: \[ \sqrt{\frac{20 \times 9^2 + 20 \times 7^2}{40}} = \sqrt{8} = 2.8 \]**4. Standard Error (SE):**- Formula: \[ SE = 2.8 \sqrt{\frac{1}{21} + \frac{1}{21}} \]- Calculation steps: \[ = 2.8 \sqrt{0.0952} = 2.8284 \times 0.3086 \]- Result: \[ = 0.8729 \]**5. Standard Error:**- Final value: \[ = 0.8729 \]This information can be used for concluding statistical significance between two sample means based on the calculated standard error and critical values.

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Description: I just need help with d,e,fi attached the work for a,b,c A researcher conducts a study to determine whether spending time outdoors improves happiness. She randomly assigns 42 participants into two groups. One group serves as the control group, while

To testH0:μ1=μ2H1:μ1>μ2 d)The t-statistics ist=M1-M2SE=8-60.8729=20.8729=2.2912

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Statistics

(d) Compute the t-statistic for this hypothesis test and describe what the t-score tells you (e) Determine whether the outdoor exposure program increased happiness, explaining your decision (f) Compute Cohen's d and describe what this number tells you in this specific study

(d) Compute the t-statistic for this hypothesis test and describe what the t-score tells you (e) Determine whether the outdoor exposure program increased happiness, explaining your decision (f) Compute Cohen's d and describe what this number tells you in this specific study

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015

1st Edition

ISBN:9781680331141

Author:HOUGHTON MIFFLIN HARCOURT

Publisher:HOUGHTON MIFFLIN HARCOURT

Chapter4: Writing Linear Equations

Section: Chapter Questions

Problem 14CR

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Concept explainers

Permutations and Combinations

If there are 5 dishes, they can be relished in any order at a time. In permutation, it should be in a particular order. In combination, the order does not matter. Take 3 letters a, b, and c. The possible ways of pairing any two letters are ab, bc, ac, ba, cb and ca. It is in a particular order. So, this can be called the permutation of a, b, and c. But if the order does not matter then ab is the same as ba. Similarly, bc is the same as cb and ac is the same as ca. Here the list has ab, bc, and ac alone. This can be called the combination of a, b, and c.

Counting Theory

The fundamental counting principle is a rule that is used to count the total number of possible outcomes in a given situation.

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Question

I just need help with d,e,f

i attached the work for a,b,c

A researcher conducts a study to determine whether spending time outdoors improves happiness. She randomly assigns 42 participants into two groups. One group serves as the control group, while the treatment group is scheduled for outdoor activities 4 days per week. The researcher measures the happiness of each participant and the data reveal the following at the conclusion of her study:

- Treatment Group: $ n = 21, M_1 = 8, SS = 9 $
- Control Group: $ n = 21, M_2 = 6, SS = 7 $

What type of design is this study (single-sample, independent measures, repeated measures)?

(a) State the null and alternative hypotheses

(b) Using an alpha level of 0.05 ($ \alpha = 0.05 $), identify the critical t-value for this hypothesis test.

(c) Compute the standard error for this statistical analysis AND describe what this number tells you specifically for this type of design

(d) Compute the t-statistic for this hypothesis test and describe what the t-score tells you

(e) Determine whether the outdoor exposure program increased happiness, explaining your decision

(f) Compute Cohen’s d and describe what this number tells you in this specific study

**Hypotheses:**
- $ H_0: M_1 = M_2 $
- $ H_1: M_1 \neq M_2 $

**1. Degree of Freedom:**
- Calculated as $ 21 + 21 - 2 = 40 $.

**2. Critical Values:**
- Alpha level ($ \alpha $) = 0.05 for a two-tailed test.
- Two tails critical value = $ \pm 2.021 $.

**3. Pooled Standard Deviation:**
- Formula:
\[
\sqrt{\frac{(21 - 1)9^2 + (21 - 1)7^2}{21 + 21 - 2}}
\]
- Calculated as:
\[
\sqrt{\frac{20 \times 9^2 + 20 \times 7^2}{40}} = \sqrt{8} = 2.8
\]

**4. Standard Error (SE):**
- Formula:
\[
SE = 2.8 \sqrt{\frac{1}{21} + \frac{1}{21}}
\]
- Calculation steps:
\[
= 2.8 \sqrt{0.0952} = 2.8284 \times 0.3086
\]
- Result:
\[
= 0.8729
\]

**5. Standard Error:**
- Final value:
\[
= 0.8729
\]

This information can be used for concluding statistical significance between two sample means based on the calculated standard error and critical values.

Expert Solution

Step 1

To test
$H_{0} : μ_{1} = μ_{2} H_{1} : μ_{1} > μ_{2}$

d)
The t-statistics is
$\begin{array}{rcl} t & = & \frac{M_{1} - M_{2}}{S E} \\ = & \frac{8 - 6}{0.8729} \\ = & \frac{2}{0.8729} \\ = & 2.2912 \end{array}$

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