The mayor of a particular city recently claimed in a press conference that he ha

Question

The mayor of a particular city recently claimed in a press conference that he had done a tremendous job of boosting the city's economy with his policies. As evidence, he claimed that the average income for an adult city resident was $62000, which was far more than the rest of the region. An editor at the newspaper believes that the mayor's claim is too high and that the average income for city residents is actually lower. To test this, a random sample of 65 adult city residents was selected. The sample average income was $54590 and the sample standard deviation was $18632. Can the newspaper editor dispute the mayor's claim? Use a 1% significance level to test. What are the correct hypotheses?

set a)

HO: mu < $62000

HA: mu > $62000

HO: mu = $62000

HA: mu < $62000

set c)

HO: mu = $62000

HA: mu > $62000

set d) HO: mu = $62000

         HA: mu "not=" $62000

What is the correct conclusion?

a)

reject Ho. The population mean income in the city is less than $62000

b)

reject Ho. There is not enough evidence to say the population mean income in the city is less than $62000

c)

fail to reject Ho. The population mean income in the city is less than $62000

d)

fail to reject Ho. There is not enough evidence to say the population mean income in the city is less than $62000

set a)

HO: mu < $62000

HA: mu > $62000

set b)

HO: mu = $62000

HA: mu < $62000

set c)

HO: mu = $62000

HA: mu > $62000

set d) HO: mu = $62000

         HA: mu "not=" $62000

What is the correct conclusion?

a)

reject Ho. The population mean income in the city is less than $62000

b)

reject Ho. There is not enough evidence to say the population mean income in the city is less than $62000

c)

fail to reject Ho. The population mean income in the city is less than $62000

d)

fail to reject Ho. There is not enough evidence to say the population mean income in the city is less than $62000

Explanation / Answer

Set Up Hypothesis
Null, H0: U=62000
Alternate, H1: U<62000
Test Statistic
Population Mean(U)=62000
Sample X(Mean)=54590
Standard Deviation(S.D)=18632
Number (n)=65
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =54590-62000/(18632/Sqrt(65))
to =-3.206
| to | =3.206
Critical Value
The Value of |t | with n-1 = 64 d.f is 2.386
We got |to| =3.206 & | t | =2.386
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Left Tail -Ha : ( P < -3.2064 ) = 0.00105
Hence Value of P0.01 > 0.00105,Here we Reject Ho

[ANSWER]

HO: mu = $62000
HA: mu < $62000

a)
   
reject Ho. The population mean income in the city is less than $62000

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