14. A factory makes pretzels. Each mega-size bag of pretzels should weigh 2.5 lbs. The manager suspects the

filling machine has been underfilling the bags. She randomly selects 27 bags and weighs them,
calculating a sample mean of 2.38 lbs. with a sample standard deviation of 0.5 lbs. She performs a
significance test using a significance level of a =0.05 and calculates a t-value of -1.247 Recommend
a strategic decision for the factory manager.
(1 point)
The manager should keep the filling machine running without changes.
The manager should stop production to set the machine to fill the bags more.
The manager should stop production to set the machine to fill the bags less.
The manager should change the size of the bag to match the amount the machine dispenses

The manager should stop production to set the machine to fill the bags more.

Since the sample mean (2.38 lbs) is less than the expected mean (2.5 lbs), it suggests that the bags are being underfilled by the filling machine. The calculated t-value of -1.247 is less than the critical t-value of -1.703 (with 26 degrees of freedom and a significance level of 0.05). This means that we fail to reject the null hypothesis that the bags are being filled at the correct weight. However, since the sample mean is significantly lower than the expected mean, the manager should take action and adjust the filling machine to fill the bags more to meet the expected weight.

To make a strategic decision based on the given information, we need to perform a hypothesis test.

Null hypothesis: The filling machine dispenses the correct amount (2.5 lbs) on average.
Alternative hypothesis: The filling machine underfills the bags (less than 2.5 lbs) on average.

Since the sample size is small (n = 27) and the population standard deviation is unknown, we'll use a t-test.

The t-value calculated by the manager is -1.247, and the significance level (α) is 0.05.

To make a decision, we compare the calculated t-value with the critical t-value. The critical t-value is determined based on the degrees of freedom (n-1) and the significance level. The critical t-value for a one-tailed t-test with 26 degrees of freedom and a significance level of 0.05 is approximately -1.706.

Since the calculated t-value (-1.247) is greater than the critical t-value (-1.706), we fail to reject the null hypothesis. This means there is not enough evidence to conclude that the filling machine underfills the bags. Therefore, the manager should keep the filling machine running without changes. Option 1: The manager should keep the filling machine running without changes is the recommended strategic decision.