Use technology to help you test the claim about the population mean, mu, at the
ID: 3207878 • Letter: U
Question
Use technology to help you test the claim about the population mean, mu, at the given level of significance, alpha, using the given sample statistics. Assume the population is normally distributed.
Claim: mugreater than>1290; alphaequals=0.03; sigmaequals=200.82. Sample statistics: x overbarxequals=1316.49, nequals=300
A random sample of 8282 eighth grade students' scores on a national mathematics assessment test has a mean score of 283283. This test result prompts a state school administrator to declare that the mean score for the state's eighth graders on this exam is more than 280280. Assume that the population standard deviation is 3434. At alphaequals=0.050.05, is there enough evidence to support the administrator's claim?
Write the claim mathematically and identify Upper H 0H0 and Upper H Subscript aHa. Choose the correct answer below.
The lengths of time (in years) it took a random sample of 32 former smokers to quit smoking permanently are listed. Assume the population standard deviation is 4.1 years. At alphaequals=0.08, is there enough evidence to reject the claim that the mean time it takes smokers to quit smoking permanently is 14 years?
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Use technology and a t-test to test the claim about the population mean mu at the given level of significance alpha using the given sample statistics. Assume the population is normally distributed.
Claim: mugreater than>78; alphaequals=0.05 Sample statistics: x overbarxequals=778.5, sequals=3.2, nequals=23
Use a t-test to test the claim about the population mean mu at the given level of significance alpha using the given sample statistics. Assume the population is normally distributed.
Claim: munot equals29; alphaequals=0.10 Sample statistics: x overbarxequals=26.5, sequals=4.8, nequals=11
Explanation / Answer
Q.
Claim: mugreater than>1290; alphaequals=0.03; sigmaequals=200.82. Sample statistics: x overbarxequals=1316.49, nequals=300
A.
Given that,
population mean(u)=1290
standard deviation, =200.82
sample mean, x =1316.49
number (n)=300
null, Ho: =1290
alternate, H1: >1290
level of significance, = 0.03
from standard normal table,right tailed z /2 =1.881
since our test is right-tailed
reject Ho, if zo > 1.881
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 1316.49-1290/(200.82/sqrt(300)
zo = 2.28473
| zo | = 2.28473
critical value
the value of |z | at los 3% is 1.881
we got |zo| =2.28473 & | z | = 1.881
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : right tail - ha : ( p > 2.28473 ) = 0.01116
hence value of p0.03 > 0.01116, here we reject Ho
ANSWERS
---------------
null, Ho: =1290
alternate, H1: >1290
test statistic: 2.28473
critical value: 1.881
decision: reject Ho
p-value: 0.01116
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