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.

Given:
- Claimed population mean: $\mu = 2000$ hours
- $\Sigma x = 127808$
- $\Sigma(\bar{x} - x)^2 = 9694.6$
- Level of significance: $\alpha = 0.01$ (1%)

Step 1: Find the sample size and sample mean

Assuming the sample size $n = 64$ (a standard assumption when $\Sigma x = 127808$ gives a round sample mean):
$$\bar{x} = \frac{\Sigma x}{n} = \frac{127808}{64} = 1997 \text{ hours}$$

Step 2: Calculate the sample standard deviation
$$s^2 = \frac{\Sigma(\bar{x} - x)^2}{n-1} = \frac{9694.6}{63} \approx 153.88$$
$$s = \sqrt{153.88} \approx 12.4 \text{ hours}$$

Step 3: State the Hypotheses

The question asks if the manufacturer is over-estimating (i.e., actual mean is less than 2000), so this is a left-tailed test:
$$H_0: \mu = 2000 \quad \text{(manufacturer's claim is correct)}$$
H_1: \mu < 2000 \quad \text{(manufacturer is over-estimating)}

Step 4: Calculate the test statistic
$$t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{1997 - 2000}{12.4/\sqrt{64}} = \frac{-3}{12.4/8} = \frac{-3}{1.55} \approx -1.935$$

Step 5: Find the critical value
For a left-tailed test at $\alpha = 0.01$ with $df = 63$:
$$t_{\text{critical}} \approx -2.387$$

Step 6: Decision
|t_{\text{calculated}}| = 1.935 < 2.387 = |t_{\text{critical}}|

Since the calculated t-value does not fall in the rejection region, we fail to reject $H_0$.

Conclusion: There is no sufficient evidence at the 1% level to conclude that the manufacturer is over-estimating the life span of the light bulbs.

Given:
- Claimed population mean: $\mu = 50$ kg
- Population standard deviation: $\sigma = 1$ kg
- Sample size: $n = 60$
- Sample mean: $\bar{x} = 49.6$ kg
- Level of significance: $\alpha = 0.05$ (assumed standard, two-tailed)

Step 1: State the Hypotheses
$$H_0: \mu = 50 \text{ kg} \quad \text{(company's claim is correct)}$$
$$H_1: \mu \neq 50 \text{ kg} \quad \text{(inspector's suspicion — two-tailed)}$$

Step 2: Choose the test
Since population standard deviation $\sigma$ is known and $n = 60$ (large sample), we use the Z-test.

Step 3: Calculate the test statistic
$$Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} = \frac{49.6 - 50}{1 / \sqrt{60}}$$

$$Z = \frac{-0.4}{1/7.746} = \frac{-0.4}{0.1291} \approx -3.098$$

Step 4: Find the critical value
For a two-tailed test at $\alpha = 0.05$:
$$Z_{\text{critical}} = \pm 1.96$$

Step 5: Decision Rule
Reject $H_0$ if |Z_{\text{calculated}}| > 1.96

|Z_{\text{calculated}}| = 3.098 > 1.96

Step 6: Conclusion
Since 3.098 > 1.96, we reject $H_0$ (Null hypothesis is accepted means the inspector's claim is supported).

The inspector is right in his suspicions. There is sufficient statistical evidence that the mean mass of the bags is significantly different from 50 kg. The company appears to be under-filling the bags.

Given:
- Population mean (average heart rate): $\mu = 72$ beats/min
- Sample size: $n = 25$
- Sample mean: $\bar{x} = 69$ beats/min
- Sample standard deviation: $s = 6.5$ beats/min
- Level of significance: $\alpha = 0.05$ (assumed)

Step 1: State the Hypotheses

Since we want to test if the aerobics programme lowered the heart rate, this is a left-tailed test:
$$H_0: \mu = 72 \quad \text{(aerobics has no effect)}$$
H_1: \mu < 72 \quad \text{(aerobics lowered heart rate)}

Step 2: Choose the test
Since population standard deviation is unknown and sample size is small ($n = 25$), we use the t-test.

Degrees of freedom: $df = n - 1 = 24$

Step 3: Calculate the test statistic
$$t = \frac{\bar{x} - \mu}{s / \sqrt{n}} = \frac{69 - 72}{6.5 / \sqrt{25}}$$

$$t = \frac{-3}{6.5 / 5} = \frac{-3}{1.3} \approx -2.308$$

Step 4: Find the critical value
For a left-tailed test at $\alpha = 0.05$ with $df = 24$:
$$t_{\text{critical}} = -1.711$$

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

t_{\text{calculated}} = -2.308 < -1.711 = t_{\text{critical}}

Step 6: Conclusion
Since -2.308 < -1.711, we reject $H_0$.

Yes, the aerobics programme was effective in significantly lowering the heart rate of the participants at the 5% level of significance. There is sufficient statistical evidence that the mean heart rate after the programme (69 beats/min) is significantly lower than the population average of 72 beats/min.