Do shoppers at the mall spend less money on average the day after Thanksgivingcompared to the day after Christmas? The 58 randomly surveyed shoppers on the dayafter Thanksgiving spent an average of $126. Their standard deviation was $35. The58 randomly surveyed shoppers on the day after Christmas spent an average of $111.Their standard deviation was $32. What can be concluded at the a = 0.01 level ofsignificance? Assuming Equal VariancesTor this study, we shoutd useSelect an answera The null and alternative hypotheses-would beHgSelect an-anower Select an answer v Select an answer v (please enter adecimal)H:Setect an answer v Select an answer VSelect an answer v (Please enter adecimalb. The test stattstT is a. The test statistic vatue -(please show your answer to 3 decimalptaces.d. The p-value =(Please show your answer to 4 decimal places.)e. The p-value is ? af. Based on this, we should Select an answer v the null hypothesis.g. Thus, the conclusion is that ...Select an answer

Answered: d. The p-value = |(Please show your…

d. The p-value = |(Please show your answer to 4 decimal places.) e. The p-value is a f. Based on this, we should Select an answer | the null hypothesis.

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Question

Do shoppers at the mall spend less money on average the day after Thanksgiving
compared to the day after Christmas? The 58 randomly surveyed shoppers on the day
after Thanksgiving spent an average of $126. Their standard deviation was $35. The
58 randomly surveyed shoppers on the day after Christmas spent an average of $111.
Their standard deviation was $32. What can be concluded at the a = 0.01 level of
significance? Assuming Equal Variances
Tor this study, we shoutd use
Select an answer
a The null and alternative hypotheses-would be
HgSelect an-anower Select an answer v Select an answer v (please enter a
decimal)
H:Setect an answer v Select an answer V
Select an answer v (Please enter a
decimal
b. The test stattstT is a
. The test statistic vatue -
(please show your answer to 3 decimal
ptaces.
d. The p-value =
(Please show your answer to 4 decimal places.)
e. The p-value is ? a
f. Based on this, we should Select an answer v the null hypothesis.
g. Thus, the conclusion is that ...
Select an answer

Expert Solution

Step 1

Given : n1=58 , n2=58 , X1-bar=126 , X2-bar=111 , s1=35 , s2=32 ,α=0.01

Since , the population standard deviations are unknown and equal.

Here , we use the two sample mean t-test.

Therefore , the degrees of freedom is given by,

df=n1+n2-2=58+58-2=114

The pooled estimate of the standard deviation is ,

$S_{p} = \sqrt{\frac{(n_{1} - 1) s_{1}^{2} + (n_{2} - 1) s_{2}^{2}}{n_{1} + n_{2} - 2}} = \sqrt{\frac{(58 - 1) 35^{2} + (58 - 1) 32^{2}}{58 + 58 - 2}} = 33.5336$

The null and alternative hypothesis is,

Ho : μ1=μ2

H1 : μ1 ˂ μ2

The test left-tailed test.

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