Is there enough evidence to conclude that the bags do not contain 5 pounds as stated, at α = 0.05? Find the 95% confidence interval of the true mean.

Steps to Solve

1. Set Up Hypotheses

   We need to establish the null and alternative hypotheses for the statistical test.

   • Null Hypothesis ($H_0$): The mean weight of the bags is 5 pounds, $H_0: \mu = 5$.
   • Alternative Hypothesis ($H_1$): The mean weight of the bags is not equal to 5 pounds, $H_1: \mu \neq 5$.

2. Calculate the Test Statistic

   We can use the formula for the test statistic $z$ for the 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 values:

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

3. Compute Value

   Let's compute the denominator:

   $$ \frac{0.7}{\sqrt{50}} \approx \frac{0.7}{7.07} \approx 0.099 $$

   Now calculate $z$:

   $$ z = \frac{4.6 - 5}{0.099} \approx \frac{-0.4}{0.099} \approx -4.04 $$

4. Determine Critical Value and Decision Rule

   For a two-tailed test with $\alpha = 0.05$, we find the critical z-values from the z-table:

   • Critical values: $z_{0.025} \approx \pm 1.96$.

   Decision rule: If $z < -1.96$ or $z > 1.96$, we reject the null hypothesis.

5. Make the Conclusion

   Since $z \approx -4.04 < -1.96$, we reject the null hypothesis, concluding there is sufficient evidence to suggest that the bags do not contain 5 pounds.

6. Calculate 95% Confidence Interval for the Mean

   The formula for the confidence interval is:

   $$ CI = \bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} $$

   Calculating the margin of error:

   $$ ME = z_{0.025} \cdot \frac{0.7}{\sqrt{50}} = 1.96 \cdot 0.099 \approx 0.194 $$

   Now calculate the confidence interval:

   $$ CI = 4.6 \pm 0.194 \implies (4.406, 4.794) $$