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# 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 \(\alpha =0.05\)?

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### Questions & Answers

**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 \(\alpha =0.05\)?

**ANSWER:**

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.