A nutrition lable for Oriental Speice states that one package of the sauce has 1

Question

A nutrition lable for Oriental Speice states that one package of the sauce has 1,190 mg of sodium. To determine if the label is accurate, the FDA randomly selects 200 packages of Oriental Spices Sause and determines the sodium content. The sample has an average of 1,167.34 mg of sodium per package with a sample standard deviation of 252.94mg.

A. Find the p-value for the test of hypothesis that the sodium content is different than the nutrition label states.

B. Is the sufficient evidence to reject the null hypothesis at a signiicance level of 0.01?

Please show the work to all of these questions.

Thanks

Explanation / Answer

Given that,
population mean(u)=1190
sample mean, x =1167.34
standard deviation, s =252.94
number (n)=200
null, Ho: =1190
alternate, H1: !=1190
level of significance, = 0.01
from standard normal table, two tailed t /2 =2.601
since our test is two-tailed
reject Ho, if to < -2.601 OR if to > 2.601
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =1167.34-1190/(252.94/sqrt(200))
to =-1.2669
| to | =1.2669
critical value
the value of |t | with n-1 = 199 d.f is 2.601
we got |to| =1.2669 & | t | =2.601
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.2669 ) = 0.2067
hence value of p0.01 < 0.2067,here we do not reject Ho
ANSWERS
---------------
null, Ho: =1190
alternate, H1: !=1190
test statistic: -1.2669
critical value: -2.601 , 2.601
decision: do not reject Ho
p-value: 0.2067

evidence that the sodium content is not different than the nutrition label states

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