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### Problem 1: Hypothesis test on calories burned by tennis players vs. basketball players

- **Given:**
  - Mean calories burned by basketball players, \(\mu_b = 650\)
  - Sample mean calories burned by tennis players, \(\bar{x}_t = 657\)
  - Standard deviation of calories burned by tennis players, \(s_t = 41.5\)
  - Sample size for tennis players, \(n_t = 37\)
  - Significance level: \(\alpha = 0.05\)

#### Hypotheses:
- Null Hypothesis (\(H_0\)): \(\mu_t = \mu_b\) (mean calories burned by tennis players equals basketball players)
- Alternative Hypothesis (\(H_a\)): \(\mu_t > \mu_b\) (mean calories burned by tennis players is more than basketball players)

#### Test Statistic:
Since the sample size is 37 (greater than 30), we can use the Z-test:
\[
Z = \frac{\bar{x}_t - \mu_b}{\frac{s_t}{\sqrt{n_t}}}
\]

#### Critical Value:
For a one-tailed test at \(\alpha = 0.05\), the critical Z-value is approximately 1.645.

#### Decision Rule:
- If \(Z > 1.645\), reject \(H_0\); otherwise, do not reject \(H_0\).

Let me calculate the Z-value for you.

### Problem 2: Comparing Customer Satisfaction Levels between Two Companies

- **Given:**
  - \(n_1 = 174\), \(n_2 = 355\)
  - \(\bar{x}_1 = 3.51\), \(\bar{x}_2 = 3.24\)
  - \(s_1 = 0.51\), \(s_2 = 0.52\)
  - Significance level: \(\alpha = 0.01\)

#### Hypotheses:
- Null Hypothesis (\(H_0\)): \(\mu_1 = \mu_2\) (mean satisfaction of Company 1 equals Company 2)
- Alternative Hypothesis (\(H_a\)): \(\mu_1 > \mu_2\) (mean satisfaction of Company 1 is higher)

#### Test Statistic:
This is a two-sample Z-test for comparing means, and the test statistic is:
\[
Z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
\]

#### Critical Value:
For a one-tailed test at \(\alpha = 0.01\), the critical Z-value is approximately 2.33.

#### Decision Rule:
- If \(Z > 2.33\), reject \(H_0\); otherwise, do not reject \(H_0\).

### Problem 3: Comparing Persistence in Learning between Boys and Girls

- **Given:**
  - Girls: \(n_g = 14\), \(\bar{x}_g = 26.9\), \(s_g = 12\)
  - Boys: \(n_b = 11\), \(\bar{x}_b = 21.3\), \(s_b = 13\)
  - Significance level: \(\alpha = 0.01\)

#### Part (a): Unequal variances (Welch’s t-test)
#### Hypotheses:
- Null Hypothesis (\(H_0\)): \(\mu_g = \mu_b\) (mean persistence for girls equals boys)
- Alternative Hypothesis (\(H_a\)): \(\mu_g \neq \mu_b\) (mean persistence for girls is different)

#### Test Statistic:
For Welch’s t-test (unequal variances):
\[
t = \frac{\bar{x}_g - \bar{x}_b}{\sqrt{\frac{s_g^2}{n_g} + \frac{s_b^2}{n_b}}}
\]

#### Degrees of Freedom (df):
\[
df = \frac{\left(\frac{s_g^2}{n_g} + \frac{s_b^2}{n_b}\right)^2}{\frac{\left(\frac{s_g^2}{n_g}\right)^2}{n_g - 1} + \frac{\left(\frac{s_b^2}{n_b}\right)^2}{n_b - 1}}
\]

#### Critical Value:
Use the t-distribution with the appropriate degrees of freedom and \(\alpha = 0.01\).

#### Part (b): Equal variances (pooled t-test)
#### Hypotheses:
- Same as part (a).

#### Test Statistic:
For equal variances (pooled t-test):
\[
t = \frac{\bar{x}_g - \bar{x}_b}{\sqrt{s_p^2 \left(\frac{1}{n_g} + \frac{1}{n_b}\right)}}
\]
Where \(s_p^2\) is the pooled variance:
\[
s_p^2 = \frac{(n_g - 1) s_g^2 + (n_b - 1) s_b^2}{n_g + n_b - 2}
\]

### Do you want me to perform the detailed calculations for all these problems?

Solve the hypothesis tests for calorie burning, customer satisfaction comparison, and persistence in learning based on the given data.

This advanced statistics problem covers hypothesis testing for calorie burning, customer satisfaction comparison, and persistence in learning. Learn how to apply Z-tests and T-tests with significance levels of 0.05 and 0.01, respectively.

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