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 will conduct a one-tailed hypothesis test.

• Null Hypothesis ($H_0$): The mean weight of the bags is equal to 5 pounds, $H_0: \mu = 5$
• Alternative Hypothesis ($H_a$): The mean weight of the bags is less than 5 pounds, $H_a: \mu < 5$

2. Determine the Test Statistic
   We will use the formula for the test statistic ($z$) for the sample mean:
   $$ z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} $$
   Where:

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

Substituting in the values:
$$ z = \frac{4.6 - 5}{0.7 / \sqrt{50}} $$

3. Calculate the Test Statistic
   Calculate the standard error first:
   $$ SE = \frac{0.7}{\sqrt{50}} \approx 0.099 \text{ (rounded)} $$
   Now substitute back into the test statistic formula:
   $$ z = \frac{-0.4}{0.099} \approx -4.04 $$

4. Determine the Critical Value
   For a significance level of $\alpha = 0.05$ in a one-tailed test, the critical value from the z-table is approximately $-1.645$.

5. Make a Decision
   Compare the calculated z-value to the critical value:

• If $z < -1.645$, we reject the null hypothesis;
• Since $-4.04 < -1.645$, we reject $H_0$.
  This means there is sufficient evidence to conclude that the bags do not contain 5 pounds.

6. Calculate the 95% Confidence Interval
   The formula for a confidence interval for the mean is:
   $$ CI = \bar{x} \pm z_{\alpha/2} \left( \frac{\sigma}{\sqrt{n}} \right) $$
   For 95% confidence, $z_{0.025} \approx 1.96$.
   Now calculate the margin of error:
   $$ ME = 1.96 \times \frac{0.7}{\sqrt{50}} \approx 0.196 $$
   Hence, the confidence interval is:
   $$ CI = 4.6 \pm 0.196 $$

7. Final Confidence Interval Calculation
   Calculating the bounds:
   $$ CI_{lower} = 4.6 - 0.196 = 4.404 $$
   $$ CI_{upper} = 4.6 + 0.196 = 4.796 $$
   So, the 95% confidence interval is $(4.404, 4.796)$.