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

Steps to Solve

1. State the Hypotheses

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

2. Determine the significance level

   The significance level ($\alpha$) is given as 0.05, which is commonly used for hypothesis testing.

3. Calculate the test statistic

   We will use the formula for the z-test statistic:

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

   where:

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

   Plugging in the values:

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

4. Compute the z-score

   To calculate the z-score, first, calculate the denominator:

   $$ \frac{0.7}{\sqrt{50}} \approx 0.09899 $$

   Now substitute this back into the z formula:

   $$ z \approx \frac{-0.4}{0.09899} \approx -4.04 $$

5. Find the critical z-values

   For a significance level of 0.05 in a two-tailed test, the critical z-values are approximately -1.96 and 1.96.

6. Make the decision for the null hypothesis

   If the calculated z-score is less than -1.96 or greater than 1.96, we reject the null hypothesis. Since $-4.04 < -1.96$, we reject $H_0$.

7. Calculate the confidence interval

   The 95% confidence interval for the true mean can be calculated using the formula:

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

   Substituting in the values we have:

   • The z-value for 95% is approximately 1.96.
   • Calculate the margin of error:

   $$ ME = z_{\frac{\alpha}{2}} \cdot \frac{\sigma}{\sqrt{n}} = 1.96 \cdot 0.09899 \approx 0.194 $$

   Now we can find the confidence interval:

   $$ \text{CI} = (4.6 - 0.194, 4.6 + 0.194) = (4.406, 4.794) $$