Resolver los problemas de estadística que involucran el cálculo de intervalos de confianza y pruebas de hipótesis.

Steps to Solve

1. Problem 1: Calculate the sample mean and standard deviation

First, calculate the sample mean ($\bar{x}$) and sample standard deviation ($s$) from the given data: 9.8, 10.2, 10.4, 9.8, 10.0, 10.2, 9.6.

$$ \bar{x} = \frac{9.8 + 10.2 + 10.4 + 9.8 + 10.0 + 10.2 + 9.6}{7} = \frac{70}{7} = 10 $$

$$ s = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n-1}} $$

$$ s = \sqrt{\frac{(9.8-10)^2 + (10.2-10)^2 + (10.4-10)^2 + (9.8-10)^2 + (10.0-10)^2 + (10.2-10)^2 + (9.6-10)^2}{7-1}} $$

$$ s = \sqrt{\frac{0.04 + 0.04 + 0.16 + 0.04 + 0 + 0.04 + 0.16}{6}} = \sqrt{\frac{0.48}{6}} = \sqrt{0.08} \approx 0.2828 $$ 2. Determine the t-value for the confidence interval

Since the sample size ($n = 7$) is small, we use the t-distribution. We want a 95% confidence interval, so $\alpha = 1 - 0.95 = 0.05$. The degrees of freedom are $df = n - 1 = 7 - 1 = 6$. Looking up the t-value for $\alpha/2 = 0.025$ and $df = 6$ in a t-table, we find $t_{0.025, 6} \approx 2.447$.

3. Calculate the margin of error

The margin of error ($E$) is calculated as:

$$ E = t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}} = 2.447 \cdot \frac{0.2828}{\sqrt{7}} \approx 2.447 \cdot \frac{0.2828}{2.6458} \approx 2.447 \cdot 0.1069 \approx 0.2616 $$

4. Calculate the confidence interval

The confidence interval is given by:

$$ \bar{x} - E < \mu < \bar{x} + E $$

$$ 10 - 0.2616 < \mu < 10 + 0.2616 $$

$$ 9.7384 < \mu < 10.2616 $$

5. Problem 2a: Calculate the margin of error

Given: $n = 50$, $\bar{x} = 174.5$, $s = 6.9$, confidence level = 98%, so $\alpha = 1 - 0.98 = 0.02$. Since $n > 30$, we can use the z-distribution. For a 98% confidence interval, we need the z-score for $\alpha/2 = 0.01$. Looking up $z_{0.01}$ in a z-table, we find $z_{0.01} \approx 2.33$.

The margin of error ($E$) is calculated as:

$$ E = z_{\alpha/2} \cdot \frac{s}{\sqrt{n}} = 2.33 \cdot \frac{6.9}{\sqrt{50}} \approx 2.33 \cdot \frac{6.9}{7.071} \approx 2.33 \cdot 0.9758 \approx 2.2736 $$ 6. Calculate the confidence interval

The confidence interval is given by:

$$ \bar{x} - E < \mu < \bar{x} + E $$

$$ 174.5 - 2.2736 < \mu < 174.5 + 2.2736 $$

$$ 172.2264 < \mu < 176.7736 $$

7. Problem 2b: State the margin of error

The margin of error is the value $E$ calculated in the previous part. We can say with 98% confidence that the possible size of our error is approximately 2.2736 cm.

8. Problem 3: Calculate the sample mean and standard deviation

First, calculate the sample mean ($\bar{x}$) and sample standard deviation ($s$) from the given data: 1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 0.99, 1.01, 1.03.

$$ \bar{x} = \frac{1.01 + 0.97 + 1.03 + 1.04 + 0.99 + 0.98 + 0.99 + 1.01 + 1.03}{9} = \frac{9.05}{9} \approx 1.0056 $$

$$ s = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n-1}} $$

$$ s = \sqrt{\frac{(1.01-1.0056)^2 +(0.97-1.0056)^2 + (1.03-1.0056)^2 + (1.04-1.0056)^2 + (0.99-1.0056)^2 + (0.98-1.0056)^2 + (0.99-1.0056)^2 + (1.01-1.0056)^2 + (1.03-1.0056)^2}{9-1}} $$

$$ s = \sqrt{\frac{0.00001936 + 0.00126736 + 0.00059536 + 0.00118336 + 0.00024336 + 0.00065536 + 0.00024336 + 0.00001936 + 0.00059536}{8}} $$

$$ s = \sqrt{\frac{0.00482224}{8}} \approx \sqrt{0.00060278} \approx 0.02455 $$ 9. Determine the t-value for the confidence interval

Since the sample size ($n = 9$) is small, we use the t-distribution. We want a 99% confidence interval, so $\alpha = 1 - 0.99 = 0.01$. The degrees of freedom are $df = n - 1 = 9 - 1 = 8$. Looking up the t-value for $\alpha/2 = 0.005$ and $df = 8$ in a t-table, we find $t_{0.005, 8} \approx 3.355$.

10. Calculate the margin of error

The margin of error ($E$) is calculated as:

$$ E = t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}} = 3.355 \cdot \frac{0.02455}{\sqrt{9}} \approx 3.355 \cdot \frac{0.02455}{3} \approx 3.355 \cdot 0.008183 \approx 0.02745 $$

11. Calculate the confidence interval

The confidence interval is given by:

$$ \bar{x} - E < \mu < \bar{x} + E $$

$$ 1.0056 - 0.02745 < \mu < 1.0056 + 0.02745 $$

$$ 0.97815 < \mu < 1.03305 $$

12. Problem 4: State the null and alternative hypotheses

Null hypothesis ($H_0$): $\mu = 40$ months Alternative hypothesis ($H_1$): $\mu < 40$ months (one-tailed test)

13. Calculate the test statistic

Given: $n = 64$, $\bar{x} = 38$, $s = 5.8$, $\mu_0 = 40$, $\alpha = 0.05$.

The test statistic (t-value) is calculated as:

$$ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{38 - 40}{5.8 / \sqrt{64}} = \frac{-2}{5.8 / 8} = \frac{-2}{0.725} \approx -2.7586 $$ 14. Determine the critical value and make a decision

Since it's a one-tailed test with $\alpha = 0.05$ and $df = n - 1 = 64 - 1 = 63$, we look up the critical t-value in a t-table. Since 63 is not on the table, we will use 60, $t_{0.05, 60} \approx -1.671$. Since our calculated t-value (-2.7586) is less than the critical t-value (-1.671), we reject the null hypothesis.

15. State the conclusion

We reject the null hypothesis. There is sufficient evidence to conclude that the average lifespan of the mice on the specified diet is less than 40 months at a 5% significance level.

16. Problem 5: State the null and alternative hypotheses

Null hypothesis ($H_0$): $\mu = 10$ liters Alternative hypothesis ($H_1$): $\mu \neq 10$ liters (two-tailed test)

17. Calculate the sample mean and standard deviation

Given data: 10.2, 9.7, 10.1, 10.3, 10.1, 9.8, 9.9, 10.4, 10.3, 9.8

$$ \bar{x} = \frac{10.2 + 9.7 + 10.1 + 10.3 + 10.1 + 9.8 + 9.9 + 10.4 + 10.3 + 9.8}{10} = \frac{100.6}{10} = 10.06 $$

$$ s = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n-1}} $$

$$ s = \sqrt{\frac{(10.2-10.06)^2 + (9.7-10.06)^2 + (10.1-10.06)^2 + (10.3-10.06)^2 + (10.1-10.06)^2 + (9.8-10.06)^2 + (9.9-10.06)^2 + (10.4-10.06)^2 + (10.3-10.06)^2 + (9.8-10.06)^2}{10-1}} $$

$$ s = \sqrt{\frac{0.0196 + 0.1296 + 0.0016 + 0.0576 + 0.0016 + 0.0676 + 0.0256 + 0.1156 + 0.0576 + 0.0676}{9}} $$

$$ s = \sqrt{\frac{0.544}{9}} \approx \sqrt{0.06044} \approx 0.2459 $$ 18. Calculate the test statistic

$$ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{10.06 - 10}{0.2459 / \sqrt{10}} = \frac{0.06}{0.2459 / 3.1623} \approx \frac{0.06}{0.07776} \approx 0.7716 $$

19. Determine the critical value and make a decision

Since it's a two-tailed test with $\alpha = 0.01$ and $df = n - 1 = 10 - 1 = 9$, we look up the critical t-value in a t-table for $\alpha/2 = 0.005$. We find $t_{0.005, 9} \approx \pm 3.250$. Since our calculated t-value (0.7716) is between -3.250 and 3.250, we fail to reject the null hypothesis.

20. State the conclusion

We fail to reject the null hypothesis. There isn't sufficient evidence to conclude that the manufacturer is wrong about the content of the oil containers at a 1% significance level.