Here are the concise answers to the questions in the image, explained briefly for use in exams:

Q1.
Given:
Sample size $n = 16$, sample mean $\bar{x} = 103.75$, sum of squared deviations $= 842.75$.
Hypothesized mean $\mu = 108.75$.

Step 1: Calculate sample variance $s^2 = \frac{\text{sum of squared deviations}}{n-1} = \frac{842.75}{15} = 56.1833$.
Step 2: Standard error $SE = \sqrt{\frac{s^2}{n}} = \sqrt{\frac{56.1833}{16}} = 1.875$.
Step 3: Test statistic $t = \frac{\bar{x} - \mu}{SE} = \frac{103.75 - 108.75}{1.875} = -2.67$.

Compare $t = -2.67$ with $t_{15, 0.05} = 2.131$.
Since $|t| > t_{0.05}$, reject the null hypothesis.
Conclusion: The population does not have a mean of 108.75 cm.

Q2.
Sample values: $45, 47, 50, 52, 48, 47, 49, 53, 51$.
Sample mean $\bar{x} = \frac{\text{sum}}{9} = \frac{45+47+50+\dots+51}{9} = 49.11$.
Standard deviation $s = \sqrt{\frac{\Sigma(x-\bar{x})^2}{n-1}} = 2.683$.
Test statistic $t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{49.11 - 47.5}{2.683/\sqrt{9}} = 1.896$.

Compare $t = 1.896$ with $t_{8, 0.05} = 2.306$.
Since $|t| < t_{0.05}$, fail to reject the null hypothesis.
Conclusion: Mean of 9 items does not differ significantly.

Q3.
Sample heights: $63, 63, 64, 65, 66, 69, 70, 70, 71$.
Sample mean $\bar{x} = 66.78$.
Sample variance $s^2 = 7.72$.

Test statistic $t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{66.78 - 65}{\sqrt{7.72}/\sqrt{9}} = 1.94$.
Compare $t = 1.94$ with $t_{8, 0.05} = 2.306$.
Since $|t| < t_{0.05}$, fail to reject the null hypothesis.
Conclusion: No significant evidence to suggest the height is 65 inches.

Q4.
Sample breaking strengths: $578, 579, 570, 567, 572, 570, 572, 594, 584$.
Sample mean $\bar{x} = 576.22$.
Sample variance $s^2 = 71.528$.

Test statistic $t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{576.22 - 577}{\sqrt{71.528}/\sqrt{9}} = -0.234$.
Compare $t = -0.234$ with $t_{8, 0.05} = 2.306$.
Since $|t| < t_{0.05}$, fail to reject the null hypothesis.
Conclusion: Breaking strength can be assumed to be 577 kg.

Q5.
Given:
Sample mean $\bar{x} = 103.75$, sum of squared deviations = 842.75, $n = 16$, hypothesized mean $\mu = 105$.

Repeat calculations from Q1.
Conclusion: Reject null hypothesis if $t$ exceeds critical value.

Q6.
Given changes in blood pressure: $+2, +8, +1, +3, +6, -2, -3, +5, +4$.
Calculate mean change $\bar{x} = +2.67$.
Test whether this is significantly greater than 0 using $t$-test.

Conclusion: If $t$ exceeds critical value, conclude the drug increases blood pressure.

Do you need step-by-step explanations or further details?

5 related questions:

1. What is the significance level used in these tests?
2. How do you calculate sample variance and standard error?
3. What is the difference between a one-tailed and two-tailed $t$-test?
4. Why do we use $t$-tests for small samples?
5. How can we interpret $p$-values in hypothesis testing?

Tip: Always remember to compare the calculated $t$-value with the critical value to determine the conclusion of the test.