9:44 АМ Thu Dec 10- 97%Click Here to begin, or go to assignments and click on midterm 3Jack is a golfer, and his actual mean drive distance last year was 260 yards. Hehas been practicing a new technique all year. He now wants to test to see if he hasmade an improvement. The distances are given below. Perform a hypothesis test atthe 10% significance level to see if his mean drive distance is now greater than last6 of 8year's average of 260.270280249260264250258259280270262272290255261259Assume that the true population of distances is normally distributed with standarddeviation 9.8.a) Calculate the sample mean. (To save you some time, the sum is 4239)b) Set up the null and alternative hypothesesc) State the significance level, a, in probability form.PreviousNextDashboard000000CalendarТo DoNotificationsInbox

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ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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9:44 АМ Thu Dec 10
- 97%
Click Here to begin, or go to assignments and click on midterm 3
Jack is a golfer, and his actual mean drive distance last year was 260 yards. He
has been practicing a new technique all year. He now wants to test to see if he has
made an improvement. The distances are given below. Perform a hypothesis test at
the 10% significance level to see if his mean drive distance is now greater than last
6 of 8
year's average of 260.
270
280
249
260
264
250
258
259
280
270
262
272
290
255
261
259
Assume that the true population of distances is normally distributed with standard
deviation 9.8.
a) Calculate the sample mean. (To save you some time, the sum is 4239)
b) Set up the null and alternative hypotheses
c) State the significance level, a, in probability form.
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Expert Solution

Step 1

Given information:

Significance level $(α) = 0.10$

Population mean, that is last year's average drive distance $(μ) = 260$

Population standard deviation $(σ) = 9.8$

Sample size $(n) = 16$

The given distances are as follows:

270	280	249	260
264	250	258	259
280	270	262	272
290	255	261	259

Step 2

Sum of distances given $(Σ X) = 4239$

The sample mean $(\bar{X})$ is calculated as follows:

$\begin{array}{rcl} \bar{X} & = & \frac{Σ X}{n} \\ = & \frac{4239}{16} \\ = & 264.9375 \end{array}$

Therefore, the sample mean is 264.9375.

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