Question

I believe the population mean number of times that adults ate fast food each week is 2.5. My data is after speaking to with 7 and found that they ate fast food 1, 4, 0, 2, 4, 0 and 3 times last week. I choose to test this hypothesis at .08 significance level.

My hypothesis is: Ho: μ + = 2.5 HA : μ ≠ 2.5 n = 7

Calculate the p-value: 0.4738

Calculate the t-stat: = 0.764

Using the above results - I need to know how many of the 7 people surveyed, how many gained weight within the last year? I am estimating 5 of the 7 gained weight and I would like to test this at .10 significance level. I am not sure if I wrote this right but I really need help with this problem. If I didn't write it correctly, I would appreciate any help doing it correctly. Thank you.

And could you give me step by step of how it should be written and solved. Thanks again.

Answer 1

Given that,
population mean(u)=2.5
sample mean, x =2
standard deviation, s =1.7321
number (n)=7
null, Ho: μ=2.5
alternate, H1: μ!=2.5
level of significance, alpha = 0.1
from standard normal table, two tailed t alpha/2 =1.943
since our test is two-tailed
reject Ho, if to < -1.943 OR if to > 1.943
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =2-2.5/(1.7321/sqrt(7))
to =-0.764
| to | =0.764
critical value
the value of |t alpha| with n-1 = 6 d.f is 1.943
we got |to| =0.764 & | t alpha | =1.943
make decision
hence value of |to | < | t alpha | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -0.7637 ) = 0.474
hence value of p0.1 < 0.474,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=2.5
alternate, H1: μ!=2.5
test statistic: -0.764
critical value: -1.943 , 1.943
decision: do not reject Ho
p-value: 0.474
we do not have enough evidence to support claim that the population mean number of times that adults ate fast food each week is 2.5

b.
Given that,
possible chances (x)=5
sample size(n)=7
success rate ( p )= x/n = 0.714
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5
alternate, H1: p!=0.5
level of significance, alpha = 0.1
from standard normal table, two tailed z alpha/2 =1.645
since our test is two-tailed
reject Ho, if zo < -1.645 OR if zo > 1.645
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.71429-0.5/(sqrt(0.25)/7)
zo =1.134
| zo | =1.134
critical value
the value of |z alpha| at los 0.1% is 1.645
we got |zo| =1.134 & | z alpha | =1.645
make decision
hence value of |zo | < | z alpha | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.13389 ) = 0.25684
hence value of p0.1 < 0.2568,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p!=0.5
test statistic: 1.134
critical value: -1.645 , 1.645
decision: do not reject Ho
p-value: 0.2568
we do not have enough evidence to support the claim that gained weight within the last year.