A realtor specializes in selling apartments in Kelowna. Its records indicate that the mean selling time is 90 days. However, it believes that because of the recent heat waves, the mean selling time is now greater than 90 days. A survey of 26 apartments sold recently revealed that the mean selling time was 94 days, with a standard deviation of 14 days.

At the 0.03 significance level, has there been an increase in selling time?

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-bar d p-bar p (Click to select) = < > (3 points)

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

Step 2.Use =0.025

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

c. So, your conclusion is thatthe mean selling time (Click to select) increased has not increased .(1 point)

d. Estimate the p-value. Is thep-value greater than the significance level? (Click to select) No Yes .(1 point)