Question

In: Statistics and Probability

An entrepreneur examines monthly sales (in $1,000s) for 40 convenience stores in Rhode Island. (You may...

An entrepreneur examines monthly sales (in $1,000s) for 40 convenience stores in Rhode Island. (You may find it useful to reference the appropriate table: z table or t table)

Excel data

Sales Sqft
140 1810
160 2500
80 1010
180 2170
140 2310
110 1320
90 1130
110 1500
130 1950
80 1010
110 1770
140 1840
140 2330
140 2490
120 1550
120 1900
210 2320
120 1700
180 2500
170 2380
160 1880
120 1780
120 1610
90 1230
140 1920
100 1260
90 1260
190 2470
130 2420
110 1550
100 1260
140 2230
100 1500
140 1970
120 1530
120 1800
110 1520
170 2210
100 1440
110 1470

a. Select the null and the alternative hypotheses in order to test whether average sales differ from $130,000.

  • H0: μ = 130,000; HA: μ ≠ 130,000

  • H0: μ ≥ 130,000; HA: μ < 130,000

  • H0: μ ≤ 130,000; HA: μ > 130,000

b-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)

b-2. Find the p-value.

  • p-value < 0.01

  • 0.01 ≤ p-value < 0.02

  • 0.02 ≤ p-value < 0.05

  • 0.05 ≤ p-value < 0.10

  • p-value ≥ 0.10

c. At α = 0.05 what is your conclusion? Do average sales differ from $130,000?

  • Reject H0; average sales differ from $130,000.

  • Reject H0; average sales do not differ from $130,000.

  • Do not reject H0; average sales differ from $130,000.

  • Do not reject H0; average sales do not differ from $130,000.

Solutions

Expert Solution

Values ( X )
140000 138062500
160000 1008062500
80000 2328062500
180000 2678062500
140000 138062500
110000 333062500
90000 1463062500
110000 333062500
130000 3062500
80000 2328062500
110000 333062500
140000 138062500
140000 138062500
140000 138062500
120000 68062500
120000 68062500
210000 6683062500
120000 68062500
180000 2678062500
170000 1743062500
160000 1008062500
120000 68062500
120000 68062500
90000 1463062500
140000 138062500
100000 798062500
90000 1463062500
190000 3813062500
130000 3062500
110000 333062500
100000 798062500
140000 138062500
100000 798062500
140000 138062500
120000 68062500
120000 68062500
110000 333062500
170000 1743062500
100000 798062500
110000 333062500
Total 5130000 37177500000





To Test :-

H0: μ = 130,000

HA: μ ≠ 130,000

Test Statistic :-


t = -0.3585


Test Criteria :-
Reject null hypothesis if


Result :- Fail to reject null hypothesis

b-2. Find the p-value.

P value

Looking for the value t = 0.3585 across n - 1 = 40 - 1 = 39 degree of freedom for two tailed

t = 0.3585 lies between value 0.000 and 0.681 respective P value is 1.00 and 0.50

P value = 0.05 ≤ p-value < 0.10

c. At α = 0.05 what is your conclusion? Do average sales differ from $130,000

Do not reject H0

Average sales do not differ from $130,000.


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