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A popular theory is that presidential candidates have an advantage if they are taller than their main opponents. Listed are heights (in centimeters) of randomly selected presidents along with the heights of their main opponents. Complete parts (a) and (b) below.

\begin{tabular}{lllllll}
\hline Height [tex]$( cm )$[/tex] of President & 193 & 170 & 166 & 180 & 198 & 180 \\
\hline Height [tex]$( cm )$[/tex] of Main Opponent & 176 & 180 & 181 & 183 & 193 & 178 \\
\hline
\end{tabular}

a. Use the sample data with a 0.01 significance level to test the claim that for the population of heights for presidents and their main opponents, the differences have a mean greater than [tex]$0 cm$[/tex].

In this example, [tex]$\mu_{ d }$[/tex] is the mean value of the differences [tex]$d$[/tex] for the population of all pairs of data, where each individual difference [tex]$d$[/tex] is defined as the president's height minus their main opponent's height.

What are the null and alternative hypotheses for the hypothesis test?

$
\begin{array}{l}
H_0: \mu_d=0 cm \\
H_1: \mu_d>0 cm
\end{array}
$

(Type integers or decimals. Do not round.)

Identify the test statistic.

[tex]$t = \square$[/tex] (Round to two decimal places as needed.)

Sagot :

GinnyAnswer
To solve this hypothesis testing problem, let's go through it step-by-step:

### Step 1: State the Hypotheses

Given the problem, we need to test the claim that the mean of the differences in heights (president's height minus opponent's height) is greater than \(0 \text{ cm}\).

The hypotheses are stated as:
- Null Hypothesis (\(H_0\)): \(\mu_d = 0 \text{ cm}\)
- Alternative Hypothesis (\(H_1\)): \(\mu_d > 0 \text{ cm}\)

### Step 2: Identify the Test Statistic

We are using the sample data:

[tex]\[ \text{Heights of Presidents (cm)}: [193, 170, 166, 180, 198, 180] \][/tex]
[tex]\[ \text{Heights of Main Opponents (cm)}: [176, 180, 181, 183, 193, 178] \][/tex]

Firstly, we find the differences between the heights of presidents and their main opponents:

[tex]\[ \text{Differences (cm)} = [193-176, 170-180, 166-181, 180-183, 198-193, 180-178] \][/tex]

[tex]\[ = [17, -10, -15, -3, 5, 2] \][/tex]

These differences form our sample.

#### Sample Mean of the Differences (\(\bar{d}\))
The sample mean difference is:

[tex]\[ \bar{d} = \frac{17 + (-10) + (-15) + (-3) + 5 + 2}{6} \approx -0.67 \][/tex]

#### Sample Standard Deviation of the Differences (\(s_d\))
The standard deviation of the differences is approximately:

[tex]\[ s_d \approx 11.40 \][/tex]

#### Sample Size (\(n\))
The number of data points is:

[tex]\[ n = 6 \][/tex]

Now we calculate the test statistic using the formula for the t-statistic:

[tex]\[ t = \frac{\bar{d} - \mu_{d_0}}{s_d / \sqrt{n}} \][/tex]

Where:
- \(\bar{d}\) is the sample mean of the differences,
- \(\mu_{d_0}\) is the hypothesized population mean difference (which is 0),
- \(s_d\) is the sample standard deviation of the differences,
- \(n\) is the sample size.

By substituting the values, we get:

[tex]\[ t = \frac{-0.67 - 0}{11.40 / \sqrt{6}} \approx -0.14 \][/tex]

### Conclusion

So, the test statistic is:

[tex]\[ t \approx -0.14 \][/tex]

Therefore, the completed solution is:

[tex]\[ t = -0.14 \][/tex]

So, in summary, the test statistic for this hypothesis test is approximately [tex]\( t = -0.14 \)[/tex].
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