Question

The mean number of shopping trips per week in a county is 2.2. In a particular neighborhood within the county, a survey based on 50 people reveals that the mean in the neighborhood is 1.85, with a standard deviation of 1.2. Test the null hypothesis that the mean in the neighborhood is no different from that in the county. State the null and alternative hypotheses, find the test statistic, compare it with the critical value, make a decision, and find the p-value. Use alpha = 0.05.

Thank you!

Answer 1

Answer:

n=50,  = 2.2

= 1.85 , s = 1.2

= 0.05

a)

null and alternative hypothesis is

Ho:   = 2.2

H1:     2.2

c)

formula for test statistics is

t = -2.062

test statistics: t = -2.062

d)

Calculate t critical value for two tailed test with = 0.05

and df = n -1 = 50 -1 = 49

using t table we get critical values as

Critical value = 2.014

Critical value = ( -​​​2.014 , 2.014 )

e)

decision rule is

Reject Ho if ( test statistics ) < ( -​​​2.014 ) or ( test statistics ) > ( ​2.014)

here, ( test statistics = -2.062 ) < ( -​​​2.014 )

Hence,

Null hypothesis is rejected.

f)

Therefore there is not sufficient evidence to support the claim that the mean shopping trips per week in the neighborhood is no different from that in the county.

g)

using excel command we get p-value as

P-Value = 0.0445