Question

The following information consists of prices (in dollars) for one sample of California cabernet sauvignon
wines that received ratings of 93 or higher in an issue of Wine Spectator and another sample of California
cabernets that received ratings of 89 or lower in the same issue. De ne population 1 as wines that received
ratings of at least 93. A sample of size n1 = 12; x1 = 110:75 and s1 = 48:744 was obtained. De ne population
2 as wines that received ratings of at most 89. A sample of size n2 = 14; x2 = 61:1429 and s2 = 24:791 was
obtained. Assume populations 1 and 2 are normally distributed with population mean 1 and 2, respectively.
Suppose H0 : 1 m 2 = 0 and Ha : 1 m 2 > 0:
1. Calculate the value of the test statistic for the di erence in mean rating.
2. Specify the distribution of the test statistic.
3. Compute the P-value and determine if we reject null hypothesis, with = 0:05
4. There are three type of con dence intervals: two-taild, one tailed with upper bound, and one tailed with
lower bound. For this hypothesis test, choose the appropriate con dence interval for the di erence in
mean rating. Let the con dence level be 95%. Does 0 belongs to this interval? Does it support to reject
null hypothesis?

Answer 1

n1 = 12; x1 = 110.75 and s1 = 48.744

n2 = 14; x2 = 61.1429 and s2 = 24.791

1)

sd= sqrt(48.744 ^2/12 + 24.791^2/14) =15.55306

z= 110.75-61.1429/ sd = 3.18954

2)

the difference is a normal distribution with mean m1-m2 and standard deviation SD defined above

3)

this is one sided test with upper bound

3)

p value = 0.00074

from R

for one sided t test critical vaule at ,05 is 1.64

as t claulated> t critical we rejec tthe null and say the difference is > 0

CI = 49.6071 -1.96* 15.55306 *sqrt(1/12+1/14), 49.6071 + 1.96*15.55306*sqrt(1/12+1/14)

=37.61476,61.59944

0 is not included

yes it supports as both ends of the CI are well above 0