Question

A consumer product testing organization uses a survey of readers to obtain customer satisfaction ratings for the nation's largest supermarkets. Each survey respondent is asked to rate a specified supermarket based on a variety of factors such as: quality of products, selection, value, checkout efficiency, service, and store layout. An overall satisfaction score summarizes the rating for each respondent with 100 meaning the respondent is completely satisfied in terms of all factors. Suppose sample data representative of independent samples of two supermarkets' customers are shown below.

Supermarket 1	Supermarket 2
n1 = 280	n2 = 300
x1 = 89	x2 = 88

(a)

Formulate the null and alternative hypotheses to test whether there is a difference between the population mean customer satisfaction scores for the two retailers. (Let μ₁ = the population mean satisfaction score for Supermarket 1's customers, and let μ₂ = the population mean satisfaction score for Supermarket 2's customers. Enter != for ≠ as needed.)

H₀:

H_a:

(b)

Assume that experience with the satisfaction rating scale indicates that a population standard deviation of 14 is a reasonable assumption for both retailers. Conduct the hypothesis test.

Calculate the test statistic. (Use μ₁ − μ₂. Round your answer to two decimal places.)

Report the p-value. (Round your answer to four decimal places.)

p-value =

At a 0.05 level of significance what is your conclusion?

Reject H₀. There is not sufficient evidence to conclude that the population mean satisfaction scores differ for the two retailers.Do not reject H₀. There is sufficient evidence to conclude that the population mean satisfaction scores differ for the two retailers.     Reject H₀. There is sufficient evidence to conclude that the population mean satisfaction scores differ for the two retailers.Do not reject H₀. There is not sufficient evidence to conclude that the population mean satisfaction scores differ for the two retailers.

(c)

Which retailer, if either, appears to have the greater customer satisfaction?

Supermarket 1 Supermarket 2     neither

Provide a 95% confidence interval for the difference between the population mean customer satisfaction scores for the two retailers. (Use x₁ − x₂.Round your answers to two decimal places.)

_______ to ________

Answer 1

Given that,
mean(x)=89
standard deviation , σ1 =14
number(n1)=280
y(mean)=88
standard deviation, σ2 =14
number(n2)=300
null, Ho: u1 = u2
alternate, H1: μ1 != u2
level of significance, alpha = 0.05
from standard normal table, two tailed z alpha/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=89-88/sqrt((196/280)+(196/300))
zo =0.86
| zo | =0.86
critical value
the value of |z alpha| at los 0.05% is 1.96
we got |zo | =0.86 & | z alpha | =1.96
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 != 0.86 ) = 0.39001
hence value of p0.05 < 0.39001,here we do not reject Ho
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(a) null, Ho: u1 = u2
   alternate, there is a difference between the population mean customer satisfaction scores for the two retailers H1: μ1 != u2
(b)
test statistic: 0.86
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.39001
(c)
CI = x1 - x2 ± Z 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)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 89-88) ±Z a/2 * Sqrt( 196/280+196/300)]
= [ (1) ± Z a/2 * Sqrt( 1.3533) ]
= [ (1) ± 1.96 * Sqrt( 1.3533) ]
= [-1.2801 , 3.2801]
both appears to be similar customer satisfaction