The mean number of shopping tr
The mean number of shopping trips per week in a county is 2.2.In a particular neighborhood within the county, a survey based on50 people reveals that the mean in the neighborhood is 1.85, with astandard deviation of 1.2. Test the null hypothesis that the meanin the neighborhood is no different from that in the county. Statethe null and alternative hypotheses, find the test statistic,compare it with the critical value, make a decision, and find thep-value. Use alpha = 0.05.
Thank you!
Answer:
Answer:
n=50, 
= 2.2
= 1.85 , s= 1.2
=0.05
a)
null andalternative hypothesis is
Ho: 
= 2.2
H1: 
2.2
c)
formulafor test statistics is
t =-2.062
test statistics: t =-2.062
d)
Calculate t criticalvalue for two tailed test with 
=0.05
and df =n -1 = 50 -1 = 49
using ttable we get critical values as
Critical value = 
2.014
Critical value = ( -2.014 , 2.014)
e)
decisionrule is
Reject Ho if ( teststatistics ) < ( -2.014 ) or ( test statistics ) > (2.014)
here, ( test statistics= -2.062 ) < ( -2.014 )
Hence,
Null hypothesis isrejected.
f)
Therefore there is notsufficient evidence to support the claim that the mean shoppingtrips per week in the neighborhood is no different from that in thecounty.
g)
using excel command we get p-value as
P-Value =0.0445
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