Question

The day after Thanksgiving is known as "Black Friday" as retailers open early on Friday morning, or sometimes late at night on Thanksgiving Day itself, to start the holiday shopping season. However, there is a current social movement afoot to allow employees to enjoy the Thanksgiving holiday by discouraging consumers from patronizing stores that open on Thanksgiving Day evening. SaMart believes that consumers will not follow the social pressure and will continue to show up in large numbers at midnight on Thanksgiving Day to start "Black Friday." Consider the following hypothesis test that measures the average number of customers waiting outside all SaMart locations to start "Black Friday": H0: μ ≥ 173 customers usually show up early on Black Friday HA: μ < 173 customers will show up (i.e. fewer) if the "social movement" occurs It's now Black Friday. At our 69 SaMart stores, there are an average of 170 customers waiting for those SaMart stores to open early. The population standard deviation is known to be 25. Use Table 1. We are going to test whether customers continue to show up early on Black Friday, or whether the "social movement" has resulted in a statistically observed change in customer behavior for SaMart.

A.) What is the critical value for the test with α = 0.10 and with α = 0.01? (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.)

Critical Value

α = 0.10=

α = 0.01=

B.) Calculate the value of the test statistic. (If there is a negative value, it should be indicated by a minus sign. Round your answer to 2 decimal places.) NOTE: the next automated question will ask you the p-value that corresponds to this test statistic. Be prepared to provide that numerical answer.

Test statistic=

C. What was the p-value that corresponds to your test statistic?; and,

At the level of α in part d. of the previous problem, what's the practical implication of this hypothesis test; that is, what does it mean to SaMart and opening late-night for Black Friday? (Write a short answer showing me you know what it means to accept or reject the null hypothesis in this SaMart question.)

Answer 1

Population standard deviation is known, so we will use z distribution

(A) z critical value at alpha = 0.10

z critical = NORMSINV(alpha)

= NORMSINV(0.10)

= -1.28

z critical value at alpha = 0.01

z critical = NORMSINV(alpha)

= NORMSINV(0.01)

= -2.33

(B) Using TI 84 calculator

press stat then tests then Ztest

press ENTER

test statisic = -1.00

(C) p value = 0.1587 (4 decimals) or 0.159(3 decimals). (select p value according to number of decimals mentioned in question)

p value is greater than 0.01 as well as 0.10 significance level, so we failed to reject the null hypothesis. Therefore, we can conclude that the mean is at least 173