Inferential Statistics — NCERT Solutions

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.