David Hughes would like to determine whether there are more units produced on the afternoon shift than on the morning shift. A sample of 63 morning-shift workers showed that the mean number of units produced was 344. A sample of 69 afternoon-shift workers showed that the mean number of units produced was 349. Assume that the population standard deviation for the number of unitsproduced on the morning shift is 16 and on the afternoon shift is 25.

At the 0.04 significance level, is the number of units produced on the afternoon shift larger?

Use these tables to determine thez-value(s)ort-value(s).

Round all z-values to 2 decimal placesand t-values to 3 decimal places.

Unless otherwise stated, report proportions and probabilitiesas decimals values (not %) and round them to 4 decimal places.

Do not round intermediate results or, if you do, round them to 5 decimal places.

Step 1. State the null hypothesis and the alternate hypothesis.

If the population parameter is negative, enter it with a "-" sign.

H₀: (Click to select) x-bar1 - x-bar2 1 - 2 p-bar1 - p-bar2 p1 - p2 d (Click to select) = < > (3 points)

H₁: same as above (Click to select) = < > same as above (1 point)

Step 2.Use =0.04

Step 3 & 4. Identify the critical value and formulate thedecision rule.

If the critical value (CV) and the test statisticare negative, enter them with a "-" sign.

If there are 2 critical values (e.g. z < -1.65 or z >1.65), select"< -CV OR > +CV" and enter only the positivevalue.

Reject H₀ if (Click to select) z t (Click to select) < > = < -CV OR > +CV .(3 points)

Step 5.Make a decision

a. The value of the test statistic is .(1 point)

b.The p-value is equal to . (1 point)

c. Based on that, your decisionis to (Click to select) reject accept H₀.(1 point)

d. So, your conclusion is that there is (Click to select) enough evidence not enough evidence that the number of units produced on the afternoon shift is larger.(1 point)