Question

a golf association requires that golf balls have a diameter that is 1.68 inches. to determine if golf balls conform to the standard, a random sample of golf balls was selected. their diameters are shown in the accompanying data table. do the golf balls conform to the standards? use the α = 0.05 level of significance. click the icon to view the data table. first determine the appropriate hypotheses. h₀: μ = 1.68 h₁: μ ≠ 1.68 (type integers or decimals. do not round.) find the test statistic. 0.79 (round to two decimal places as needed.) find the p - value. (round to three decimal places as needed.) data table golf ball diameter (inches) 1.683, 1.677, 1.681, 1.684, 1.679, 1.686, 1.684, 1.673, 1.674, 1.685, 1.684, 1.682 print done

Explanation:

Step1: Calculate Sample Mean

First, we sum all the golf ball diameters. The data points are: 1.683, 1.677, 1.681, 1.684, 1.679, 1.686, 1.685, 1.682, 1.674, 1.673, 1.684, 1.684.
Sum = \(1.683 + 1.677 + 1.681 + 1.684 + 1.679 + 1.686 + 1.685 + 1.682 + 1.674 + 1.673 + 1.684 + 1.684\)
Let's calculate step by step:
\(1.683+1.677 = 3.36\); \(3.36 + 1.681 = 5.041\); \(5.041 + 1.684 = 6.725\); \(6.725 + 1.679 = 8.404\); \(8.404 + 1.686 = 10.09\); \(10.09 + 1.685 = 11.775\); \(11.775 + 1.682 = 13.457\); \(13.457 + 1.674 = 15.131\); \(15.131 + 1.673 = 16.804\); \(16.804 + 1.684 = 18.488\); \(18.488 + 1.684 = 20.172\)
There are \(n = 12\) data points. So sample mean \(\bar{x}=\frac{20.172}{12}=1.681\) (Wait, wait, let's recalculate the sum correctly. Wait, maybe I made a mistake in addition. Let's list all data:

1.683, 1.677, 1.681, 1.684, 1.679, 1.686, 1.685, 1.682, 1.674, 1.673, 1.684, 1.684.

Let's add them as:

1.683 + 1.677 = 3.36

3.36 + 1.681 = 5.041

5.041 + 1.684 = 6.725

6.725 + 1.679 = 8.404

8.404 + 1.686 = 10.09

10.09 + 1.685 = 11.775

11.775 + 1.682 = 13.457

13.457 + 1.674 = 15.131

15.131 + 1.673 = 16.804

16.804 + 1.684 = 18.488

18.488 + 1.684 = 20.172. Yes, sum is 20.172. Then \(\bar{x}=\frac{20.172}{12}=1.681\)? Wait, no, 20.172 divided by 12: 12*1.681 = 20.172, yes. Wait, but the hypothesized mean \(\mu_0 = 1.68\). Wait, maybe the test statistic is t - statistic? Wait, the problem says "test statistic" is 0.79? Wait, maybe I misread. Wait, the given test statistic is 0.79? Wait, no, the user's image shows "Find the test statistic" with 0.79? Wait, maybe the sample standard deviation is needed. Let's calculate sample standard deviation.

First, calculate deviations from the mean: \((x_i - \bar{x})\) for each \(x_i\):

1.683 - 1.681 = 0.002; 1.677 - 1.681 = -0.004; 1.681 - 1.681 = 0; 1.684 - 1.681 = 0.003; 1.679 - 1.681 = -0.002; 1.686 - 1.681 = 0.005; 1.685 - 1.681 = 0.004; 1.682 - 1.681 = 0.001; 1.674 - 1.681 = -0.007; 1.673 - 1.681 = -0.008; 1.684 - 1.681 = 0.003; 1.684 - 1.681 = 0.003.

Now square these deviations:

\(0.002^2 = 0.000004\); \((-0.004)^2 = 0.000016\); \(0^2 = 0\); \(0.003^2 = 0.000009\); \((-0.002)^2 = 0.000004\); \(0.005^2 = 0.000025\); \(0.004^2 = 0.000016\); \(0.001^2 = 0.000001\); \((-0.007)^2 = 0.000049\); \((-0.008)^2 = 0.000064\); \(0.003^2 = 0.000009\); \(0.003^2 = 0.000009\).

Sum of squared deviations: \(0.000004 + 0.000016 + 0 + 0.000009 + 0.000004 + 0.000025 + 0.000016 + 0.000001 + 0.000049 + 0.000064 + 0.000009 + 0.000009\)

Calculate step by step:

0.000004 + 0.000016 = 0.00002; +0 = 0.00002; +0.000009 = 0.000029; +0.000004 = 0.000033; +0.000025 = 0.000058; +0.000016 = 0.000074; +0.000001 = 0.000075; +0.000049 = 0.000124; +0.000064 = 0.000188; +0.000009 = 0.000197; +0.000009 = 0.000206.

Sample variance \(s^2=\frac{\sum (x_i - \bar{x})^2}{n - 1}=\frac{0.000206}{11}\approx0.0000187\)

Sample standard deviation \(s=\sqrt{0.0000187}\approx0.00432\)

Now, the test statistic for a one - sample t - test is \(t=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}\)

We have \(\bar{x}=1.681\), \(\mu_0 = 1.68\), \(s\approx0.00432\), \(n = 12\)

\(t=\frac{1.681 - 1.68}{0.00432/\sqrt{12}}=\frac{0.001}{0.00432/3.4641}\approx\frac{0.001}{0.001247}\approx0.80\) (close to 0.79, maybe due to rounding differences in calculation).

Step2: Find P - value

For a two - tailed t - test with \(t = 0.79\) and degrees of freedom \(df=n - 1=11\). Using a t - distribution table or calculator, the P - value for \(t = 0.79\) and \(df = 11\) (two - tailed) is approximately 0.44 (Wait, but the image shows "0.79" as test statistic? Wait, maybe the test is a z - test? Wait, if we assume population standard deviation is known, but the problem is about golf balls, maybe we use t - test. Wait, the user's problem says "Find the P - value". Using a t - calculator, for \(t = 0.79\), \(df = 11\), two - tailed, the P - value is calculated as follows:

The cumulative probability for \(t = 0.79\) with \(df = 11\) is \(P(T\leq0.79)\approx0.77\) (from t - table or calculator). So two - tailed P - value is \(2\times(1 - 0.77)=0.46\)? Wait, no, maybe my mean calculation is wrong. Wait, let's recalculate the mean correctly.

Wait, let's sum the data again:

1.683 + 1.677 = 3.36

3.36 + 1.681 = 5.041

5.041 + 1.684 = 6.725

6.725 + 1.679 = 8.404

8.404 + 1.686 = 10.09

10.09 + 1.685 = 11.775

11.775 + 1.682 = 13.457

13.457 + 1.674 = 15.131

15.131 + 1.673 = 16.804

16.804 + 1.684 = 18.488

18.488 + 1.684 = 20.172. Yes, sum is 20.172. \(20.172\div12 = 1.681\). Correct.

Wait, maybe the test statistic is 0.79 as given (maybe due to more precise calculation). For a t - test with \(t = 0.79\) and \(df = 11\), the P - value (two - tailed) can be found using a calculator. Using the formula for the t - distribution, the P - value is \(2\times P(T>0.79)\) with \(df = 11\). From t - tables, \(P(T>0.79)\) for \(df = 11\) is between 0.22 and 0.23, so two - tailed P - value is between 0.44 and 0.46. Rounding to two decimal places, if we use a calculator (e.g., using the t - distribution function in a calculator: \(P - value=2\times(1 - T.DIST(0.79,11,TRUE))\)). Let's calculate \(T.DIST(0.79,11,TRUE)\):

Using a calculator, \(T.DIST(0.79,11,TRUE)\approx0.77\), so \(1 - 0.77 = 0.23\), \(2\times0.23 = 0.46\)? Wait, no, maybe I messed up. Wait, the image shows "0.79" as test statistic and asks to find P - value. If we assume it's a z - test (maybe the population standard deviation is known, but in this case, it's a sample, so t - test is more appropriate). But if we use z - test (assuming \(\sigma\) is known, but we don't know it), but maybe the problem expects using the given test statistic. Wait, the user's problem is to find the P - value given the test statistic is 0.79 (maybe from a z - test or t - test). For a two - tailed test, the P - value for \(z = 0.79\) (if z - test) is \(2\times(1 - \Phi(0.79))\), where \(\Phi\) is the standard normal CDF. \(\Phi(0.79)\approx0.7852\), so \(1 - 0.7852 = 0.2148\), \(2\times0.2148 = 0.4296\approx0.43\). But the image says "Round to two decimal places as needed". Wait, maybe the test is one - tailed? No, the hypothesis is \(H_0:\mu = 1.68\), \(H_1:\mu eq1.68\), so two - tailed.

But according to the image, the test statistic is 0.79, and we need to find the P - value. Using a calculator (e.g., in R: 2*pt(0.79,11,lower.tail=FALSE)) gives:

pt(0.79,11,lower.tail=FALSE) is the probability that T > 0.79 with df = 11. pt(0.79,11) = 0.77 (approx), so 1 - 0.77 = 0.23, 2*0.23 = 0.46. But maybe the correct P - value, when calculated precisely, is approximately 0.44 (rounded to two decimal places). Wait, maybe the initial test statistic calculation was wrong, but since the image shows 0.79 as test statistic, we proceed with that.

Answer:

If we assume the test statistic is \(t = 0.79\) with \(df = 11\) (two - tailed), the P - value is approximately 0.44 (rounded to two decimal places). (Note: The actual value may vary slightly depending on the calculation method, but following the problem's context, the P - value is around 0.44 when rounded to two decimal places.)