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Golf Robots. Serious golfers and golf equipment companies sometimes use golf equipment testing labs to obtain precise infor mation about particular club heads, club shafts, and golf balls. One golfer requested information about the Jazz Fat Cat 5-iron from Golf Laboratories, Inc. The company tested the club by using a robot to hit a Titleist NXT Tour ball six times with a head velocity of $85$miles per hour. The golfer wanted a club that, on average, would hit the ball more than $180$yards at that club speed. The total yards each ball traveled was as follows.

a. At the $5\%$significance level, do the data provide sufficient evi dence to conclude that the club does what the golfer wants? (Note: The sample mean and sample standard deviation of the data are $182.7$yards and $2.7$yards, respectively.)

b. Repeat part (a) for a test at the $1\%$significance level.

Step by step solution

01

Step 1. Given information is: 

Let denote the mean yards traveled by a golf ball.

The null and alternative hypothesis are:

$$\begin{gathered} {\mathrm{H}}_0:\mathrm{渭}=180\mathrm{yards} \\ {\mathrm{H}}_{\mathrm{a}}:\mathrm{渭}>180\mathrm{yards} \\ \mathrm{伪}=0.05 \\ \mathrm{n}=6 \\ \stackrel{炉}{\mathrm{x}}=182.7 \\ \mathrm{s}=2.7 \end{gathered}$$

02

Part (a) Step 1. Calculating P-value 

$$\begin{gathered} \mathrm{Test}\mathrm{static},\mathrm{t}=\frac{\stackrel{炉}{\mathrm{x}}-{\mathrm{渭}}_0}{{\displaystyle \raisebox{1ex}{$\mathrm{s}$}\!\left/ \!\raisebox{-1ex}{$\sqrt{\mathrm{n}}$}\right.}}...(*) \\ \mathrm{Under}\mathrm{the}\mathrm{assumption}\mathrm{that}{\mathrm{H}}_0\mathrm{is}\mathrm{true},\mathrm{t}\mathrm{follows}\mathrm{t}\mathrm{distribution}\mathrm{with}\mathrm{df}=6-1=5 \\ \mathrm{Observed}\mathrm{value}\mathrm{of}\mathrm{test}\mathrm{static}, \\ {\mathrm{t}}_0=\frac{182.7-180}{{\displaystyle \raisebox{1ex}{$2.7$}\!\left/ \!\raisebox{-1ex}{$\sqrt{6}$}\right.}}=2.45 \\ \mathrm{Since}\mathrm{the}\mathrm{given}\mathrm{hypothesis}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{tailed}\mathrm{test},\mathrm{P}\mathrm{value}\mathrm{is}\mathrm{given}\mathrm{by}: \\ \mathrm{P}-\mathrm{value}=\mathrm{P}(\mathrm{t}鈮�{\mathrm{t}}_0),\mathrm{where}\mathrm{t}~{\mathrm{t}}_5 \\ =\mathrm{P}(\mathrm{t}鈮�2.45) \\ =1-\mathrm{P}(\mathrm{t}<2.45) \\ =0.028968 \end{gathered}$$

03

Part (a) Step 2. Calculating P using MINITAB 

$$\begin{gathered} \mathrm{The}\mathrm{probability},\mathrm{P}(\mathrm{t}<2.45)\mathrm{is}\mathrm{calculated}\mathrm{using}\mathrm{MINITAB}\mathrm{in}\mathrm{the}\mathrm{follwing}\mathrm{way}: \\ \mathrm{Step}1:\mathrm{Press}\mathrm{the}\mathrm{Calc}\mathrm{menu};\mathrm{Highlight}\mathrm{the}\text{'}\mathrm{Probability}\mathrm{Distributions}\text{'}. \\ \mathrm{Step}2:\mathrm{Press}\mathrm{t}...; \\ \mathrm{Step}3:\mathrm{Tick}\mathrm{Cumulative}\mathrm{Probability}\mathrm{and}\mathrm{enter}\mathrm{the}\mathrm{df} \\ \mathrm{Degrees}\mathrm{of}\mathrm{freedom}:5 \\ \mathrm{Step}4:\mathrm{Tick}\mathrm{Input}\mathrm{constant}\mathrm{and}\mathrm{enter}\mathrm{the}\mathrm{value}2.45 \\ \mathrm{Input}\mathrm{constant}:2.45 \\ \mathrm{Step}5:\mathrm{Press}\mathrm{Ok} \\ \mathrm{Now},\mathrm{P}=0.029<\mathrm{伪}=0.05 \\ \mathrm{Therefore},\mathrm{at}5\%\mathrm{level}\mathrm{of}\mathrm{significance}\mathrm{we}\mathrm{reject}\mathrm{the}\mathrm{null}\mathrm{hypothesis},{\mathrm{H}}_0:\mathrm{渭}=180\mathrm{yards} \end{gathered}$$

04

Part (b) Step 1. For 1% significance level

$$\begin{gathered} \mathrm{Significance}\mathrm{Level}=1\%\mathrm{or}0.01 \\ \mathrm{Now},\mathrm{P}=0.029>\mathrm{伪}=0.01 \\ \mathrm{Therefore},\mathrm{at}1\%\mathrm{level}\mathrm{of}\mathrm{significance}\mathrm{we}\mathrm{do}\mathrm{not}\mathrm{reject}\mathrm{the}\mathrm{null}\mathrm{hypothesis},{\mathrm{H}}_0:\mathrm{渭}=180\mathrm{yards} \end{gathered}$$

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