A recent study found that 70 children who watched a commercial for potato chips
ID: 3223799 • Letter: A
Question
A recent study found that 70 children who watched a commercial for potato chips featuring a celebrity endorser ate a mean of 44 grams of potato chips as compared to a mean of 30 grams for 60 children who watched a commercial for an alternative food snack. Suppose that the sample standard deviation for the children who watched the celebrity-endorsed commercial was 22.3 grams and the sample standard deviation for the children who watched the alternative food snack commercial was 13.11 grams. Complete parts (a) through (c) below. A: Assuming that the population variances arc equal and alpha-0.05 is there evidence that the mean amount of potato chips eaten was significantly higher for the children who watched the celebrity-endorsed commercial? Let population I be the weights of potato chips eaten by children who watched the celebrity-endorsed commercial and let population 2 be the weights of potato chips eaten by children who watched the alternative food snack commercial. What are the correct null and alternative hypotheses? B. What is the Test Statistic? Corresponding p-value? Assuming that the population variances arc equal, construct a 95% confidence interval estimate of the difference mu 1 - mu 2 between the mean amount of potato chips eaten by the children who watched the celebrity-endorsed commercial and children who watched the alternative food snack commercial. s lessthanorequalto mu 1 = mu 2 lessthanorequalto Please demonstrate how you got the corresponding p value as well as the T stat. What formula would be used in excel?Explanation / Answer
a.
Given that,
mean(x)=44
standard deviation , s.d1=22.3
number(n1)=70
y(mean)=30
standard deviation, s.d2 =13.11
number(n2)=60
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, a = 0.05
from standard normal table,right tailed t a/2 =1.657
since our test is right-tailed
reject Ho, if to > 1.657
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (69*497.29 + 59*171.8721) / (130- 2 )
s^2 = 347.2927
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=44-30/sqrt((347.2927( 1 /70+ 1/60 ))
to=14/3.2786
to=4.2701
| to | =4.2701
critical value
the value of |t a| with (n1+n2-2) i.e 128 d.f is 1.657
we got |to| = 4.2701 & | t a | = 1.657
make decision
hence value of | to | > | t a| and here we reject Ho
p-value: right tail -ha : ( p > 4.2701 ) = 0.00002
hence value of p0.05 > 0.00002,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 4.2701
critical value: 1.657
decision: reject Ho
p-value: 0.00002
b.
CI = x1 - x2 ± t a/2 * Sqrt(S^2(1/n1+1/n2))
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=44
Standard deviation( sd1 )=22.3
Sample Size(n1)=70
Mean(x2)=30
Standard deviation( sd2 )=13.11
Sample Size(n2)=60
S^2 = (69*497.29 + 59*171.8721) / (130- 2 )
S^2 = 347.29268672
CI = [ ( 44-30) ± t a/2 * 18.63579048 Sqrt( 1/70+1/60)]
= [ (14) ± t a/2 * 18.63579048 * Sqrt( 0.03095238) ]
= [ (14) ± 2.00099538 * 18.63579048 * Sqrt( 0.03095238) ]
= [ (7.43943963 , 20.56056037 ]