Super Sneaker Company is evaluating two different materials, A and B, to be used

Question

Super Sneaker Company is evaluating two different materials, A and B, to be used to construct the soles of their new active shoe targeted to city high school students in Canada. While material B costs less than material A, the company suspects that mean wear for material B is greater than mean wear for material A. Two study designs were initially developed to test this suspicion. In both designs, Halifax was chosen as a representative city of the targeted market. In Study Design 1, 14 high school students were drawn at random from the Halifax School District database. After obtaining their shoe sizes, the company manufactured 14 pairs of shoes, each pair with one shoe having a sole constructed from material A and the other shoe, a sole constructed from material B.

After 3 months, the amount of wear in each shoe was recorded in standardized units as follows: A17.1314.1712.7715.55 13.4013.1714.5514.4316.3412.9613.6316.8014.7814.20 B15.4815.8014.7714.3016.0415.0314.0415.3715.3414.6013.7813.1417.1814.51 A null hypothesis for the test is: Correct: Incorrect-14.76 Incorrect x-X Incorrect 17.18 [1 pt(s)] ou are correct. Your receipt no. is 156-3706 Previous Tries An alternative hypothesis is: Incorrect x-Taco correct:-,- D Incorrect 1 pt(s)] are correct. Your recelt no.is156-7553 .Previous Ties [3 pt(s)] To test whether mean wear for material B is greater than mean wear for material A, calculate a test statistic. 0.8557 Submit Answer Incorrect. Tries 2 /3 Previous Tries Which of the statistical tables should you use? Incorrect -distribution Incorrect z-distribution Correct: t-distribution [1 pt(s)] You are correct. Your receipt no. is 156-3262 Previous Tries 1 pt(s) How many degrees of freedom are associated with the test statistic? Submit AnsWer Tries o/1 Is this a 1-sided or a 2-sided test? Correct: 1-sided Incorrect 2-sided You are correct. Your receipt no. is 156-7381Previous Tries Which interval in the table contains the p-value for the test? p-value s 0.005 0.005

Explanation / Answer

Given that,
mean(x)=14.5629
standard deviation , s.d1=1.4149
number(n1)=14
y(mean)=14.9557
standard deviation, s.d2 =1.0282
number(n2)=14
null, Ho: u1 > u2
alternate, H1: u1 < u2
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.771
since our test is left-tailed
reject Ho, if to < -1.771
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =14.5629-14.9557/sqrt((2.00194/14)+(1.0572/14))
to =-0.84
| to | =0.84
critical value
the value of |t | with min (n1-1, n2-1) i.e 13 d.f is 1.771
we got |to| = 0.8403 & | t | = 1.771
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:left tail - Ha : ( p < -0.8403 ) = 0.20796
hence value of p0.05 < 0.20796,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 - u2 = 0
alternate, H1: u1 - u2 >0
test statistic: -0.8403
critical value: -1.771
decision: do not reject Ho
p-value: 0.20796

DIRECT METHOD
given that,
mean(x)=14.5629
standard deviation , s.d1=1.4149
sample size, n1=14
y(mean)=14.9557
standard deviation, s.d2 =1.0282
sample size,n2 =14
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 14.5629-14.9557) ± t a/2 * sqrt((2.002/14)+(1.057/14)]
= [ (-0.393) ± t a/2 * 0.467]
= [-1.402 , 0.617]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-1.402 , 0.617] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion

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