Sugar is packed in 5-pound bags. An inspector suspects the bags may not contain 5 pounds. A sample of 50 bags produces a mean of 4.6 pounds and a standard deviation of 0.7 pound. a... Sugar is packed in 5-pound bags. An inspector suspects the bags may not contain 5 pounds. A sample of 50 bags produces a mean of 4.6 pounds and a standard deviation of 0.7 pound. a) Is there enough evidence to conclude that the bags do not contain 5 pounds as stated, at alpha = 0.05? b) Find the 95% confidence interval of the true mean. Read more

Steps to Solve

1. State the Hypotheses

   The null hypothesis $H_0$: The mean weight of the bags is 5 pounds, $\mu = 5$.
   The alternative hypothesis $H_a$: The mean weight of the bags is not 5 pounds, $\mu \neq 5$.

2. Calculate the Test Statistic

   Use the formula for the test statistic (z-score) for a sample mean:

   $$ z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} $$

   where:

   • $\bar{x} = 4.6$ (sample mean)
   • $\mu_0 = 5$ (hypothesized mean)
   • $\sigma = 0.7$ (standard deviation)
   • $n = 50$ (sample size)

   Plugging in the numbers:

   $$ z = \frac{4.6 - 5}{\frac{0.7}{\sqrt{50}}} $$

3. Calculate the Z-score

   First, calculate the denominator:

   $$ \frac{0.7}{\sqrt{50}} \approx 0.099 $$

   Now substitute back into the z-score formula:

   $$ z = \frac{-0.4}{0.099} \approx -4.04 $$

4. Determine the Critical Value

   For a two-tailed test with $\alpha = 0.05$, the critical z-values are approximately $\pm 1.96$.

5. Decision Rule

   If the absolute value of the calculated z-score is greater than the critical value, we reject the null hypothesis.

6. Make a Conclusion

   Since $|-4.04| > 1.96$, we reject the null hypothesis. Therefore, there is enough evidence to conclude that the bags do not contain 5 pounds.

7. Calculate the Confidence Interval

   The formula for the confidence interval is:

   $$ \bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} $$

   Using the critical value $z_{\alpha/2} \approx 1.96$:

   • Calculate the margin of error:

   $$ E = 1.96 \times \frac{0.7}{\sqrt{50}} \approx 0.196 $$

   • The confidence interval is then:

   $$ 4.6 \pm 0.196 $$

   Which gives:

   $$ (4.404, 4.796) $$