According to a survey, people in a certain country ate an average of 215 meals i

Question

According to a survey, people in a certain country ate an average of 215 meals in restaurants in 2001. The data in the accompanying table show the number of meals eaten in restaurants as determined from a random sample of people in this country in 2009. Complete parts (a) through (d) below. | Click the icon to view the data table. a. Using 0.05, test the hypothesis that the number of meals eaten at restaurants by people in this country has not changed since 2001. Determine the null and altemative hypotheses. Choose the correct answer below OA. HOP:215 and H1:>215 B. Ho =215 and H1 #215 OC. Ho : #215and H1.-215 ( D. Ho : 2215 and H1 :

Explanation / Answer

Given that,
population mean(u)=215
sample mean, x =205.05
standard deviation, s =84.2537
number (n)=20
null, Ho: =215
alternate, H1: !=215
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.093
since our test is two-tailed
reject Ho, if to < -2.093 OR if to > 2.093
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =205.05-215/(84.2537/sqrt(20))
to =-0.5281
| to | =0.5281
critical value
the value of |t | with n-1 = 19 d.f is 2.093
we got |to| =0.5281 & | t | =2.093
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -0.5281 ) = 0.6035
hence value of p0.05 < 0.6035,here we do not reject Ho
ANSWERS
---------------
a. null, Ho: =215 , alternate, H1: !=215
b. test statistic: -0.53
c. Option B: decision: do not reject Ho, data do not provide suffcient evidence to conclude
that average number of meals changed since 2011

d. option D

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