Questions: Suppose a professional golfing association requires that the standard deviation of the diameter of a golf ball be less than 0.003 inch. Determine whether these randomly selected golf balls conform to this requirement at the α=0.01 level of significance. Assume that the population is normally distributed. Click the icon to view the chi-square distribution table. 1.681 1.682 1.681 1.679 1.677 1.684 1.682 1.678 1.677 1.682 1.681 1.676 What are the correct hypotheses for this test? H0: σ=0.003 versus H1: σ<0.003 (Type integers or decimals. Do not round.) Find the sample standard deviation. s= (Round to five decimal places as needed.)

Solution Steps

The sample mean \( \mu \) is calculated as follows:

\[ \mu = \frac{\sum x_i}{n} = \frac{20.16}{12} = 1.68 \]

The sample variance \( s^2 \) is calculated using the formula:

\[ s^2 = \frac{\sum (x_i - \mu)^2}{n-1} = 1 \times 10^{-5} \]

Thus, the sample standard deviation \( s \) is:

\[ s = \sqrt{s^2} = \sqrt{1 \times 10^{-5}} = 0.00252 \]

The test statistic \( \chi^2 \) is calculated using the formula:

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

Substituting the values:

\[ \chi^2 = \frac{(12 - 1) \cdot 1 \times 10^{-5}}{(0.003)^2} = 12.2222 \]

To find the P-value, we calculate the cumulative distribution function (CDF) of the chi-square distribution for the test statistic \( \chi^2(11) = 12.2222 \). The corresponding P-value is:

\[ P = 0.6528 \]

The critical value for a left-tailed test at the significance level \( \alpha = 0.01 \) with \( n-1 = 11 \) degrees of freedom is:

\[ \text{Critical Value} = 3.0535 \]

Since the P-value \( 0.6528 \) is greater than the significance level \( \alpha = 0.01 \), we fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the standard deviation of the golf balls is less than \( 0.003 \) inch.

Final Answer