A researcher wishes to determine the difference in two population means.To do this, she randomly samples 9 items from each population and computes a 90% confidence interval.The sample from the first population produces a mean of 780 with a standard deviation of 240.The sample from the second population produces a mean of 890 with a standard deviation of 280.Assume that the values are normally distributed in each population.The t value used for this is ___.
A)1.860
B)1.734
C)1.746
D)1.337
E)2.342

Question

Quizplus · Accepted Answer

We need to use the formula for the confidence interval of the difference in means:

CI = (X1 - X2) ± t(α/2) * SE

where:
X1 = sample mean from population 1
X2 = sample mean from population 2
t(α/2) = t-value for the given confidence level and degrees of freedom
SE = standard error of the difference in means

Degrees of freedom = (n1 - 1) + (n2 - 1) = 8 + 8 = 16
From the given data:
X1 = 780, s1 = 240, n1 = 9
X2 = 890, s2 = 280, n2 = 9
alpha = 1 - 0.90 = 0.10, so α/2 = 0.05
The standard error of the difference in means is:

SE = sqrt[s1^2/n1 + s2^2/n2] = sqrt[(240^2/9) + (280^2/9)] = 180.80

Using a t-distribution table or calculator with 16 degrees of freedom and a 0.05 level of significance, the t-value is 1.746.

Therefore, the final confidence interval for the difference in means is:

CI = (780 - 890) ± 1.746 * 180.80
CI = -110 ± 316.16
CI = (-426.16, 206.16)

The t-value used for this is 1.746 (choice C).

A Researcher Wishes to Determine the Difference in Two Population