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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: $s_p^2 = \frac{(n_g - 1) s_g^2 + (n_b - 1) s_b^2}{n_g + n_b - 2}$

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