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defects 2 2 1 3 3 4 2 2 2 2 1 2 1 3 1 2 2 5 6 1 0 1 1 1 5 1 4 2 4 0 2 1 2 4 0 1 3 2 3 1 1 5 3 2 3 4 6 1 4 6 2. 3/4 points |Previous Answers My No A sanitation supervisor is interested in testing to see if the mean amount of garbage per bin is different from 50. In a random sample of 36 bins, the sample mean amount was 49.32 pounds and the sample standard deviation was 3.7 pounds. Conduct the appropriate hypothesis test using a 0.05 level of significance b) What is the test statistic? Give your answer to four decimal places. 1.103 c) What is the P-value for the test? Give your answer to four decimal places. 0.2776

Explanation / Answer

Q2.

Given that,
population mean(u)=50
sample mean, x =49.32
standard deviation, s =3.7
number (n)=36
null, Ho: =50
alternate, H1: !=50
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.0301
since our test is two-tailed
reject Ho, if to < -2.0301 OR if to > 2.0301
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =49.32-50/(3.7/sqrt(36))
to =-1.103
---------------
test statistic: -1.103
p-value: 0.2777

Q1.

Given that,
population mean(u)=1.78
sample mean, x =2.4
standard deviation, s =1.59079
number (n)=50
null, Ho: =1.78
alternate, H1: >1.78
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.6766
since our test is right-tailed
reject Ho, if to > 1.6766
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =2.4-1.78/(1.59079/sqrt(50))
to =2.756
ANSWERS
---------------
test statistic: 2.756
p-value: 0.0041

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