# Consider the following computer output. (a) Fill in the

Chapter 10, Problem 14E

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QUESTION:

Consider the following computer output.

Two-Sample T-Test and C1

$$\begin{array}{lllll} \text { Sample } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } \\ 1 & 12 & 10.94 & 1.26 & 0.36 \\ 2 & 16 & 12.15 & 1.99 & 0.5 \end{array}$$

Difference = $$mu\left( 1 \right)=mu\left( 2 \right)$$

Estimate for difference: -1.210

95% CI for difference: $$\left( -2.560,0.140 \right)$$

T-test of difference =0 (vs not =) :

T-value = ? p-value = ? DF = ?

Both use Pooled StDev = ?

(a) Fill in the missing values. Is this a one-sided or a two-sided test? Use lower and upper bounds for the p -value.

(b) What are your conclusions if $$\alpha =0.05$$? What if $$\alpha =0.01$$?

(c) This test was done assuming that the two population variances were equal. Does this seem reasonable?

(d) Suppose that the hypothesis had been $${{H}_{0}}:{{\mu }_{1}}={{\mu }_{2}}$$ versus $${{H}_{0}}:{{\mu }_{1}}<{{\mu }_{2}}$$. What would your conclusions be if $$\alpha =0.05$$?

QUESTION:

Consider the following computer output.

Two-Sample T-Test and C1

$$\begin{array}{lllll} \text { Sample } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } \\ 1 & 12 & 10.94 & 1.26 & 0.36 \\ 2 & 16 & 12.15 & 1.99 & 0.5 \end{array}$$

Difference = $$mu\left( 1 \right)=mu\left( 2 \right)$$

Estimate for difference: -1.210

95% CI for difference: $$\left( -2.560,0.140 \right)$$

T-test of difference =0 (vs not =) :

T-value = ? p-value = ? DF = ?

Both use Pooled StDev = ?

(a) Fill in the missing values. Is this a one-sided or a two-sided test? Use lower and upper bounds for the p -value.

(b) What are your conclusions if $$\alpha =0.05$$? What if $$\alpha =0.01$$?

(c) This test was done assuming that the two population variances were equal. Does this seem reasonable?

(d) Suppose that the hypothesis had been $${{H}_{0}}:{{\mu }_{1}}={{\mu }_{2}}$$ versus $${{H}_{0}}:{{\mu }_{1}}<{{\mu }_{2}}$$. What would your conclusions be if $$\alpha =0.05$$?

Step 1 of 4

(a) The missing values can be calculated using the information provided.

Two sample t-test for equal variances:

$${{H}_{0}}:{{\mu }_{1}}-{{\mu }_{2}}=0 \\$$

$${{H}_{A}}:{{\mu }_{1}}-{{\mu }_{2}}\ne 0$$

The Pooled standard deviation is:

$${{S}_{p}}=\sqrt{\dfrac{\left( {{n}_{1}}-1 \right){{S}_{1}}^{2}+\left( {{n}_{2}}-1 \right){{S}_{2}}^{2}}{{{n}_{1}}+{{n}_{2}}-2}} \\$$

$$=\sqrt{\dfrac{12\times \left( {{1.26}^{2}} \right)+16\times \left( {{1.99}^{2}} \right)}{12+16-2}} \\$$

$$=1.7194$$

The T-value can be calculated as the difference estimate divided by the standard error of the difference:

$$t=\dfrac{{{{\bar{x}}}_{1}}-{{{\bar{x}}}_{2}}}{{{S}_{P}}\sqrt{\left( \dfrac{1}{{{n}_{1}}}+\dfrac{1}{{{n}_{2}}} \right)}} \\$$

$$=\dfrac{10.94-12.15}{1.7194\sqrt{\left( \dfrac{1}{12}+\dfrac{1}{16} \right)}} \\$$

$$=-1.8427$$

Degree of freedom is:

$$df={{n}_{1}}+{{n}_{2}}-2 \\$$

$$=12+16-2 \\$$

$$=26$$

To find the p-value, you would need to consult a t-distribution table or use statistical software. The p-value represents the probability of observing a T-value as extreme as or more extreme than the calculated T-value, assuming the null hypothesis is true. The p-value is 0.0768 and is lies between 0.1 and 0.05. It is a Two-tailed test.