Questions: State H0 and Ha in words and in symbols. Then determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed. Explain your reasoning. Sketch a normal sampling distribution and shade the area for the P -value. A golf analyst claims that the standard deviation of the 18 -hole scores for a golfer is at most 3.1 strokes State the null hypothesis in words and in symbols. Choose the correct answer. A. The null hypothesis expressed in words is, "the standard deviation of the 18 -hole scores for a golfer is equal to 3.1 strokes." The null hypothesis is expressed symbolically as, "H0: σ=3.1." B. The null hypothesis expressed in words is, "the standard deviation of the 18 -hole scores for a golfer is at most 3.1 strokes." The null hypothesis is expressed symbolically as, "H0: σ ≤ 3.1." C. The null hypothesis expressed in words is, "the standard deviation of the 18 -hole scores for a golfer is at least 3.1 strokes." The null hypothesis is expressed symbolically as, "H0: σ ≥ 3.1." D. The null hypothesis expressed in words is, "the standard deviation of the 18 -hole scores for a golfer is not 3.1 strokes." The null hypothesis is expressed symbolically as, "H σ ≠ 3.1."

State H0 and Ha in words and in symbols. Then determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed. Explain your reasoning. Sketch a normal sampling distribution and shade the area for the P -value. A golf analyst claims that the standard deviation of the 18 -hole scores for a golfer is at most 3.1 strokes State the null hypothesis in words and in symbols. Choose the correct answer. A. The null hypothesis expressed in words is, "the standard deviation of the 18 -hole scores for a golfer is equal to 3.1 strokes." The null hypothesis is expressed symbolically as, "H0: σ=3.1." B. The null hypothesis expressed in words is, "the standard deviation of the 18 -hole scores for a golfer is at most 3.1 strokes." The null hypothesis is expressed symbolically as, "H0: σ ≤ 3.1." C. The null hypothesis expressed in words is, "the standard deviation of the 18 -hole scores for a golfer is at least 3.1 strokes." The null hypothesis is expressed symbolically as, "H0: σ ≥ 3.1." D. The null hypothesis expressed in words is, "the standard deviation of the 18 -hole scores for a golfer is not 3.1 strokes." The null hypothesis is expressed symbolically as, "H σ ≠ 3.1."
Transcript text: State $\mathrm{H}_{0}$ and $\mathrm{H}_{\mathrm{a}}$ in words and in symbols. Then determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed. Explain your reasoning. Sketch a normal sampling distribution and shade the area for the P -value. A golf analyst claims that the standard deviation of the 18 -hole scores for a golfer is at most 3.1 strokes State the null hypothesis in words and in symbols. Choose the correct answer. A. The null hypothesis expressed in words is, "the standard deviation of the 18 -hole scores for a golfer is equal to 3.1 strokes." The null hypothesis is expressed symbolically as, $" H_{0}: \sigma=3.1$." B. The null hypothesis expressed in words is, "the standard deviation of the 18 -hole scores for a golfer is at most 3.1 strokes." The null hypothesis is expressed symbolically as, $" H_{0}: \sigma \leq 3.1$. C. The null hypothesis expressed in words is, "the standard deviation of the 18 -hole scores for a golfer is at least 3.1 strokes " The null hypothesis is expressed symbolically as, " $H_{0}: \sigma \geq 3.1$." D. The null hypothesis expressed in words is, "the standard deviation of the 18 -hole scores for a golfer is not 3.1 strokes." The null hypothesis is expressed symbolically as, "H $\quad \sigma \neq 3.1$."
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Solution

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Solution Steps

Step 1: State the Hypotheses

The null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)) are defined as follows:

  • Null Hypothesis: \(H_0: \sigma \leq 3.1\)
  • Alternative Hypothesis: \(H_a: \sigma > 3.1\)
Step 2: Determine the Type of Test

Since the alternative hypothesis tests whether the standard deviation is greater than 3.1, this is a right-tailed test.

Step 3: Calculate the Test Statistic

The test statistic for the chi-square test is calculated using the formula:

\[ \chi^2 = \frac{(n-1) s^2}{\sigma_0^2} \]

Substituting the values:

\[ \chi^2 = \frac{(30 - 1) \cdot 10.0}{3.1^2} = \frac{29 \cdot 10.0}{9.61} = 30.1769 \]

Step 4: Determine the P-value

To find the P-value, we calculate the cumulative distribution function (CDF) of the chi-square distribution for the test statistic:

\[ P = P(\chi^2(29) > 30.1769) = 0.4052 \]

Step 5: Critical Value

The critical value for a right-tailed test at a significance level of \(\alpha = 0.05\) with 29 degrees of freedom is:

\[ \text{Critical Value} = 42.557 \]

Final Answer

  • Test Statistic: \( \chi^2 = 30.1769 \)
  • P-value: \( P = 0.4052 \)
  • Critical Value: \( 42.557 \)

The answer is boxed as follows:

\[ \boxed{H_0: \sigma \leq 3.1 \text{ and } H_a: \sigma > 3.1} \]

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