e. Conclude. IV. The price of a popular tennis racket at a national chain store

Question

e. Conclude. IV. The price of a popular tennis racket at a national chain store is $179. Portia bought 10 of the same racket at an online auction site for the following prices: 155, 179, 175, 175, 161, 155, 179, 175, 175, 161 Assuming that the auction prices of rackets are normally distributed, determine whether there is sufficient evidence in the sample, at the 5% level of significance, to conclude that the average price of the racket is less than $179 if purchased at an online auction. Note that x = 169 and s 9.80 a. State the null and alternative hypotheses. b. Determine an appropriate test statistic. c. Determine rejection region. d. Make a statistical decision e. Conclude.

Explanation / Answer

IV.

Given that,
population mean(u)=179
sample mean, x =169
standard deviation, s =9.8
number (n)=10
null, Ho: =179
alternate, H1: <179
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.833
since our test is left-tailed
reject Ho, if to < -1.833
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =169-179/(9.8/sqrt(10))
to =-3.227
| to | =3.227
critical value
the value of |t | with n-1 = 9 d.f is 1.833
we got |to| =3.227 & | t | =1.833
make decision
hence value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < -3.2268 ) = 0.00519
hence value of p0.05 > 0.00519,here we reject Ho
ANSWERS
---------------
null, Ho: =179
alternate, H1: <179
test statistic: -3.227
critical value: -1.833
decision: reject Ho
p-value: 0.00519

we have enough evidence to support the claim

TRADITIONAL METHOD
given that,
sample mean, x =169
standard deviation, s =9.8
sample size, n =10
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 9.8/ sqrt ( 10) )
= 3.099
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table,right tailed value of |t /2| with n-1 = 9 d.f is 1.833
margin of error = 1.833 * 3.099
= 5.681
III.
CI = x ± margin of error
confidence interval = [ 169 ± 5.681 ]
= [ 163.319 , 174.681 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =169
standard deviation, s =9.8
sample size, n =10
level of significance, = 0.05
from standard normal table,right tailed value of |t /2| with n-1 = 9 d.f is 1.833
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 169 ± Z a/2 ( 9.8/ Sqrt ( 10) ]
= [ 169-(1.833 * 3.099) , 169+(1.833 * 3.099) ]
= [ 163.319 , 174.681 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 163.319 , 174.681 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

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