Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

The average take-out order size for Ashoka Curry House restaurant is shown. Assu

ID: 3063584 • Letter: T

Question

The average take-out order size for Ashoka Curry House restaurant is shown. Assuming equal variances, at = 0.05, is there a significant difference in the order sizes?

(a) Choose the appropriate hypotheses. Assume 1 is the average order size on Friday night and 2 is the average order size on Saturday night.

a. H0: 1 2 = 0 vs H1: 1 2 0

b. H1: 1 2 0vs H0: 1 2 = 0

(b) Specify the decision rule. (A negative value should be indicated by a minus sign. Round your answers to 3 decimal places.)

Reject the null hypothesis if tcalc > ----- or tcalc < ---------.

(c) Find the test statistic tcalc. (A negative value should be indicated by a minus sign. Do not round your intermediate values and round your final answer to 4 decimal places.)

tcalc -----------

(d-1) Make a decision.

We (Click to select)reject . --------??? or do not reject-------????? the null hypothesis.

(d-2) State your conclusion.

We (Click to select)cancannot conclude that there is a significant difference in the order sizes.

(e-1) Use Excel to find the p-value. (Round your answer to 4 decimal places.)

p-value ----------------

(e-2) Interpret the p-value.

We (Click to select)reject---------????? or . do not reject--------????? the null hypothesis.

Please answer everything, all of this is one question only.

  Customer Order Size   Statistic Friday Night Saturday Night   Mean order size = 21.55 = 25.26   Standard deviation s1 = 4.64 s2 = 6.61   Number of orders n1 = 14 n2 = 20

Explanation / Answer

a)
H0: 1 2 = 0 vs H1: 1 2 0

b)

Reject the null hypothesis if tcalc > 2.0369 or tcalc < -2.0369.

c)

t = ( x1 - x2) / sqrt(s1^2/n1 + s2^2/n2)
= ( 21.55 - 25.26)/ sqrt(4.64^2/14 + 6.61^2/20)
= -1.9229

d)
We do not reject the null hypothesis

d2)

we cannot conclude that there is significant difference in order sizes

e)
using formula in excel =T.DIST(ABS(-1.923), 32,FALSE)

P value = 0.0651

e2)
we do not reject the null hypothesis

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote