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 = 20Explanation / 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
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