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. Define population 1 as wines that received ratings of at least 93. A sample of size n1 = 12, x¯1 = 110.75 and s1 = 48.744 was obtained. Define population 2 as wines that received ratings of at most 89. A sample of size n2 = 14, x¯2 = 61.1429 and s2 = 24.791 was obtained. Assume populations 1 and 2 are normally distributed with a population mean µ1 and µ2, respectively. Suppose H0 : µ1 ? µ2 = 0 and Ha : µ1 ? µ2 > 0. 1. Calculate the value of the test statistic for the difference in mean rating. 2. Specify the distribution of the test statistic. 3. Compute the P-value and determine if we reject the null hypothesis, with ? = 0.05 4. There are three type of confidence intervals: two-tailed, one-tailed with upper bound, and one-tailed lower bound. For this hypothesis test, choose the appropriate confidence interval for the difference in mean rating. Let the confidence level be 95%. Does 0 belong to this interval? Does it support to reject the null hypothesis?

Answer 1

Please see the below formula

so we calculate SE by putting in the values as

SE = sqrt((48.74^2)/12 + (24.79^2)/14) = 15.51

the stat is calculated as

t = (110.75-61.1429)/15.51 = 3.19

now the df is calculated as

v = ((((48.74^2)/12 + (24.79^2)/14))^2) / (((48.74^2)/12)^2)/11 + ((24.79^2)/14)^2)/13)) = 148.23

so we check the t critical value which is 1.655 for 1 tail

as the t stat is greater than t critical hence we reject the null hypothesis in favor of alternate hypotheiss

The confidence interval is given as

putting in the values for t cl = 1.655

and mean values we get

as we are interested in the "greater than part" of the difference in the mean values

Hence it becomes a 1 tail with upper bound confidence interval

(110.75 - 61.1429) +- 1.655*15.51

we get

23.93 and 75.27

hence the confidence interval does not contain 0 , so we can confidenntly reject the null hypothesis in favor of alternate hypothesis