Question

Twenty children are selected for a study on daily soda and milk consumption. The differences in consumption (Soda minus Milk) have a mean of 20 ml with a stand dev of 33 ml.(2-sided test)
Suppose we use the critical value method to conduct the test. We reject H0 if:
a. t ≥ 2.093 or t ≤ -2.093
b. t ≤ -2.093
c. t ≥ 1.729 or t ≤ -1.729
d. t ≥ 1.729
e. t ≥ 2.093
f. t ≤ -1.729

If the ME is used to construct a confidence interval we:
A) fail to reject H0 at the 5% level of significance since the value 0 is not contained in the 95% confidence interval
B) fail to reject H0 at the 5% level of significance since the value 0 is contained in the 95% confidence interval
C) reject H0 at the 5% level of significance since the value 20 is contained in the 95% confidence interval
D) reject H0 at the 5% level of significance since the value 0 is not contained in the 95% confidence interval
E)fail to reject H0 at the 5% level of significance since the value 20 is contained in the 95% confidence interval

Answer 1

Answer on Question #84852 – Math – Statistics and Probability

Twenty children are selected for a study on daily soda and milk consumption. The differences in consumption (Soda minus Milk) have a mean of 20 ml with a stand dev of 33 ml.(2-sided test)

Question

Suppose we use the critical value method to conduct the test. We reject H0 if:

a. $t \geq 2.093$ or $t \leq -2.093$

b. $t \leq -2.093$

c. $t \geq 1.729$ or $t \leq -1.729$

d. $t \geq 1.729$

e. $t \geq 2.093$

f. $t \leq -1.729$

Solution

For two tailed test: $t_{crit} = t_{0.025,19} = \pm 2.093$.

a. $t \geq 2.093$ or $t \leq -2.093$

Question

If the ME is used to construct a confidence interval we:

A) fail to reject H0 at the 5% level of significance since the value 0 is not contained in the 95% confidence interval

B) fail to reject H0 at the 5% level of significance since the value 0 is contained in the 95% confidence interval

C) reject H0 at the 5% level of significance since the value 20 is contained in the 95% confidence interval

D) reject H0 at the 5% level of significance since the value 0 is not contained in the 95% confidence interval

E) fail to reject H0 at the 5% level of significance since the value 20 is contained in the 95% confidence interval.

Solution

$$ME = t_{0.025,n-1} \frac{s}{\sqrt{n}} = 2.093 \frac{33}{\sqrt{20}} = 15.444.$$ $$95\% CI = (20 - 15.444, 20 + 15.44) = (4.556, 35.444).$$

D) reject H0 at the 5% level of significance since the value 0 is not contained in the 95% confidence interval.

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