Inferential Statistics

Given: A survey about daily mobile use by students uses random selection of twenty students from a school.

Concept: A sampling method is unbiased if every member of the population has an equal chance of being selected. It is biased if certain members are more likely to be selected than others.

Analysis:
In the given statement, students are selected randomly from the school. Random selection ensures that every student has an equal probability of being chosen, so no particular group is favoured or excluded.

Conclusion: The statement represents an Unbiased sampling method, because random sampling gives every student an equal chance of selection, thereby eliminating systematic bias.

Part (i):

Given: $\alpha = 0.01$, degrees of freedom $= 22$, left-tailed test.

Concept: For a left-tailed test, the critical value is negative. We look up the t-table for $\alpha = 0.01$ (one tail) with d.f. $= 22$.

From t-table: The t-value for $\alpha = 0.01$ (one-tailed) with d.f. $= 22$ is $2.508$.

Since it is a left-tailed test, the critical value is:
$$t_{\text{critical}} = -2.508$$

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Part (ii):

Given: $\alpha = 0.10$, degrees of freedom $= 18$, two-tailed test.

Concept: For a two-tailed test, the significance level is split equally into both tails, so each tail has $\alpha/2 = 0.05$. We look up the t-table for $\alpha/2 = 0.05$ with d.f. $= 18$.

From t-table: The t-value for $\alpha/2 = 0.05$ (one-tailed) with d.f. $= 18$ is $1.734$.

Since it is a two-tailed test, the critical values are:
$$t_{\text{critical}} = +1.734 \quad \text{and} \quad t_{\text{critical}} = -1.734$$

Given: A 95% confidence interval for the population mean is $(100,\ 300)$.

Concept: A confidence interval gives a range of plausible values for the population parameter. A 95% confidence interval means that if we were to repeat the sampling process 100 times, approximately 95 of those intervals would contain the true population mean.

Interpretation:

We are 95% confident that the true population mean $\mu$ lies between 100 and 300.

In other words:
$$100 < \mu < 300$$

This does not mean there is a 95% probability that the population mean falls in this specific interval (the population mean is a fixed value). Rather, it means the method used to construct this interval captures the true mean 95% of the time in repeated sampling.

Practical meaning: If this experiment were conducted 100 times, about 95 of the resulting confidence intervals would contain the true population mean. We are reasonably confident (with 5% chance of error) that the population mean is between 100 and 300.

Given:
- Population mean (original shoe): $\mu = 15$ months
- Sample size: $n = 30$
- Sample mean: $\bar{x} = 17$ months
- Sample standard deviation: $s = 5.5$ months
- Level of significance: $\alpha = 0.05$ (two-tailed)

Step 1: State the Hypotheses
$$H_0: \mu = 15 \quad \text{(No improvement; new shoe lasts the same)}$$
$$H_1: \mu \neq 15 \quad \text{(New shoe lasts differently — two-tailed)}$$

Step 2: Choose the appropriate test
Since the population standard deviation is unknown and we use the sample standard deviation, we use the t-test.

Degrees of freedom: $df = n - 1 = 30 - 1 = 29$

Step 3: Calculate the test statistic
$$t = \frac{\bar{x} - \mu}{s / \sqrt{n}} = \frac{17 - 15}{5.5 / \sqrt{30}}$$

$$t = \frac{2}{5.5 / 5.477} = \frac{2}{1.004} \approx 1.99$$

Step 4: Find the critical value
For a two-tailed test with $\alpha = 0.05$ and $df = 29$:
$$t_{\text{critical}} = \pm 2.045$$

Step 5: Decision Rule
Reject $H_0$ if $|t_{\text{calculated}}| > t_{\text{critical}}$

$$|t_{\text{calculated}}| = 1.99 < 2.045 = t_{\text{critical}}$$

Step 6: Conclusion
Since $1.99 < 2.045$, we fail to reject $H_0$ (Null hypothesis is accepted).

The designer's claim is NOT sufficiently supported at the 5% level of significance by the two-tailed test. The evidence is not strong enough to conclude that the new shoe lasts significantly longer than the original.