Question

ustomers at Costco spend an average of $130 per trip (The Wall Street Journal, October 6, 2010). One of Costco's rivals would like to determine whether its customers spend more per trip. A survey of the receipts of 25 customers found that the sample mean was $135.25. Assume that the population standard deviation is $10.50 and that spending follows a normal distribution. Use the .05 level of significance. Calculate the critical value

Answer 1

Answer:

we conclude that there is significant evidence to conclude that Costco rivals spends more per trip

Step-by-step explanation:

H0 : μ = 130

H1 : μ > 130

Sample size, n = 25

Test statistic :

(xbar - μ) ÷ (s/sqrt(n))

(135.25 - 130) ÷ (10.50/sqrt(25))

Test statistic :

5.25÷2.1 = 2.5

The Pvalue from test statistic using the Pvalue calculator :

T score = 2.5 ; df = 25 - 1 = 24 ; α = 0.05

The Pvalue = 0.009827

The Tcritical value at df = 25 - 1 = 24 ; α = 0.05 equals 1.711

Tstatistic > Tcritical ; reject the Null ;

2.5 > 1.711 ; We reject H0 ;

Pvalue < α ;

0.009827 < 0.05 ; We reject the H0.

Hence, we conclude that there is significant evidence to conclude that Costco rivals spends more per trip