Question

Problem 1 (5 points. The manager of Fred's Grocery is making a study of the amount of money customers spend in the store. A sample of 10 weekday morning shoppers and 15 Saturday morning shoppers revealed the following amounts spent : Weekday Saturday $18.88 $11.54 Average of Weekdays: 50.347 24.33 24.76 Average of Saturday: 57.36 27.26 35.78 Standard Deviation of Weekday: 24.26 35.79 36.87 Standard Deviation of Saturday: 22.34 42.31 46.78 53.77 56.04 62.94 58.45 73.59 60.98 76.51 62.67 88.09 66.89 71.65 75.61 81.87 B4.71 85.80 Owner Fred Snead contends that Saturday morning shoppers spend more than weekday shoppers. Use the Student t Test and a .05 significance level. a) What are the null and alternate hypotheses? What is the Decision Rule? What is the test statistic? d) What is the decision? e) What does the decision inter? f) What is the estimated P-value?

Answer 1

(a)

H0:Null Hypothesis: (Saturday morning shoppers do not spend more than weekday shoppers)

μι 2 μ2

HA: Alternative Hypothesis: (Saturday morning shoppers spend more than weekday shoppers) (Claim)

μι < μ2

(b)

$\alpha$ = 0.05

ndf = 10 + 15 - 2 = 23

One Tail Left Side Test

From Table, critical value of t = - 1.7139

Decision Rule:
Reject H0: If t < - 1.7139

(c)

Pooled Standard Deviation is given by:

sp = V (nl - 1)s12 + (n2 – 1) s22 nl+ n2 - 2

(10 - 1) × 24.262 + (15 – 1) 22.342 = 23.1103 10 + 15 - 2

Test Statistic is given by:

50.347 – 57.36 23.1103 x (1/10 +1/15) 03 X (1/10 + 1/15) = -0.7433

(d)

Since calculated value of t = - 0.7433 is greater than critical value of t= - 1.7139, the difference is not significant. Fail to reject null hypothesis.

(e)

The decision infers that the data do not support the claim that Saturday morning shoppers spend more than weekday shoppers.

(f)

t score = - 0.7433

ndf = 23

One Tail Left Side Test

By Technology, p - value= 0.2324