Last week, a sports supply company manufactured a batch of a few thousand bowlin
ID: 3361554 • Letter: L
Question
Last week, a sports supply company manufactured a batch of a few thousand bowling balls that are all labeled as having a weight of 20 pounds. As a consumer, you would like to test if theses bowling balls really weigh 20 pounds on average at significance level alpha = 0.1 . For a random sample of size 7, the sample mean weight was 18.8 and the sample variance was 0.0513. Assume that the weight of the bowling balls are normally distributed. (a) State the null and alternative hypotheses. (b) Construct the rejection region. (c) Calculate the test statistic. (d) Draw a conclusion. (e) Find the (range of) p-value of this test. Draw a conclusion based on this p-value. (f) Calculate the 90% confidence interval for the population mean weight.
I only really need through (c) done, I know how to do the rest, but I keep getting a weird answer for (c) and I'm not sure why. Thank you!
Explanation / Answer
PART A.
Given that,
population mean(u)=20
sample mean, x =18.8
standard deviation, s =0.2265
number (n)=7
null, Ho: >20
alternate, H1: <20
level of significance, = 0.1
from standard normal table,left tailed t /2 =1.44
since our test is left-tailed
reject Ho, if to < -1.44
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =18.8-20/(0.2265/sqrt(7))
to =-14.0172
| to | =14.0172
critical value
the value of |t | with n-1 = 6 d.f is 1.44
we got |to| =14.0172 & | t | =1.44
make decision
hence value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < -14.0172 ) = 0
hence value of p0.1 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: =20
alternate, H1: <20
test statistic: -14.0172
critical value: -1.44
decision: reject Ho
p-value: 0
the don't weigh 20
PART B.
TRADITIONAL METHOD
given that,
sample mean, x =18.8
standard deviation, s =0.2265
sample size, n =7
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.2265/ sqrt ( 7) )
= 0.086
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 6 d.f is 1.943
margin of error = 1.943 * 0.086
= 0.166
III.
CI = x ± margin of error
confidence interval = [ 18.8 ± 0.166 ]
= [ 18.634 , 18.966 ]
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DIRECT METHOD
given that,
sample mean, x =18.8
standard deviation, s =0.2265
sample size, n =7
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 6 d.f is 1.943
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 = [ 18.8 ± t a/2 ( 0.2265/ Sqrt ( 7) ]
= [ 18.8-(1.943 * 0.086) , 18.8+(1.943 * 0.086) ]
= [ 18.634 , 18.966 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 18.634 , 18.966 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
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