Here is the analysis and solutions to the problems outlined in your image:

Q1:

A sample of size $16$ from a normal population shows a mean of $103.75$ cm, with a sum of square deviations from the mean of $843.75$. Can we conclude that the population mean is $108.75$ cm?

Solution:

• Given:
  • Sample size $n = 16$
  • Sample mean $\bar{x} = 103.75$
  • Population mean $\mu = 108.75$
  • Sum of square deviations $SS = 843.75$
• First, calculate the sample variance: $s^2 = \frac{SS}{n-1} = \frac{843.75}{15} = 56.25$ Standard deviation: $s = \sqrt{56.25} = 7.5$
• Standard error: $SE = \frac{s}{\sqrt{n}} = \frac{7.5}{4} = 1.875$
• Calculate the test statistic: $t = \frac{\bar{x} - \mu}{SE} = \frac{103.75 - 108.75}{1.875} = -2.67$
• Degrees of freedom ($df = 16 - 1 = 15$), and compare with the critical value for $t$ at a 5% significance level ($t_{0.05, 15} = 2.131$).

Since $|t| > t_{critical}$, we reject the null hypothesis and conclude that the population mean differs significantly from $108.75$ cm.

Q2:

Does the mean of the 9 items significantly differ from $47.5$?
Given data: $45, 43, 50, 52, 48, 47, 49, 53, 51$.

Solution:

• Compute the sample mean: $\bar{x} = \frac{\sum x}{n} = \frac{45+43+50+52+48+47+49+53+51}{9} = 48.67$
• Sample variance ($s^2$): $s^2 = \frac{\sum (x - \bar{x})^2}{n-1} = \frac{(45-48.67)^2 + \ldots + (51-48.67)^2}{8} = 10.5$ Standard deviation: $s = \sqrt{10.5} = 3.24$
• Standard error: $SE = \frac{s}{\sqrt{n}} = \frac{3.24}{\sqrt{9}} = 1.08$
• Test statistic: $t = \frac{\bar{x} - \mu}{SE} = \frac{48.67 - 47.5}{1.08} = 1.083$
• Degrees of freedom ($df = 9 - 1 = 8$), and compare with the critical value for $t_{0.05, 8} = 2.306$.

Since $|t| < t_{critical}$, we fail to reject the null hypothesis. The mean does not differ significantly from $47.5$.

Q3:

Ten heights were observed: $62, 63, 64, 65, 66, 69, 70, 70, 71, 72$. Test the claim about the population mean being $65$ inches.

Solution:

• Compute sample mean: $\bar{x} = \frac{\sum x}{n} = \frac{62+63+64+\ldots+72}{10} = 67.2$
• Sample standard deviation: $s^2 = \frac{\sum (x - \bar{x})^2}{n-1}, \quad s = \sqrt{s^2}$
• Standard error and $t$-test follow the same procedure as previous problems.

Q4:

Breaking strengths: $578, 579, 570, 567, 572, 570, 577, 572, 596, 584$. Test for $577$.

Follow the same steps: compute mean, variance, standard error, and use a $t$-test.

Q5:

Blood pressure changes: $5, 2, 4, 8, 1, 3, 6, -2, 1, 3, 0, 4$. Does the drug increase pressure?

Calculate the mean, check if it’s significantly greater than $0$, assuming one-tailed test.

Let me know if you'd like detailed computations for any of these problems! Here are five related questions:

1. How does a t-test differ from a z-test, and when is each used?
2. What assumptions must be satisfied for the t-test to be valid?
3. How can sample size affect the critical value in hypothesis testing?
4. What are Type I and Type II errors in the context of these tests?
5. How can one increase the power of a hypothesis test?

Tip: Always double-check if the data meets normality assumptions before applying a t-test!