Question

Q4. 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 ?? 2 = 0 and Ha : 1 ?? 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

1.

Standard error of the mean is,

SE = sqrt[ (s1²/n1) + (s2²/n2) ]

= sqrt[ (48.744²/12) + (24.791²/14) ]

= 15.553

Test statistic, t = Differene in means / SE

= (110.75 - 61.1429) / 15.553

= 3.1896

2.

Degree of freedom for the sampling distribution is,

DF =  (s1²/n1 + s2²/n2)² / { [ (s1² / n1)² / (n1 - 1) ] + [ (s2² / n2)² / (n2 - 1) ] }

= (48.744²/12 + 24.791²/14)² / { [ (48.744² / 12)² / (12 - 1) ] + [ (24.791² / 14)² / (14 - 1) ] }

= 16 (Rounding to nearest integer)

The distribution of the test statistic is t distribution with degree of freedom 16.

3.

P-value = P[t > 3.1896] = 0.0028

As, P-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that there is significant evidence that differences between mean rating of populations 1 and 2 is greater than 0.

4.

As, the alternative hypothesis Ha has a greater than (> ) sign, the confidence interval to be used is one tailed with lower bound.

The value of t at DF = 16 and confidence level 95% is, 1.746

The lower bound of the 95% confidence level is

(Difference in mean rating) - Standard error * t

= (110.75 - 61.1429) - 15.553 * 1.746

= 22.4516

As, the lower bound of the 95% confidence level is greater than 0, the confidence interval does not contain 0 and thus it rejects null hypothesis.