Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given golfers the ability to hit the ball so far. Designers, therefore, have proposed that new golf courses need to be built expecting that the average golfer can hit the ball more than 230 yards on average. A random sample of 177 golfers show that their mean driving distance is 230.7 yards, with a standard deviation of 41.8.1. Set up the null and alternative hypotheses to test for the designers belief.
a. H0 : µ ≥ 230 versus Ha : µ < 230
b. H0 : µ = 230.7 versus Ha : µ ≠ 230.7
c. H0 : µ ≥ 230.7 versus Ha : µ < 230.7
d. H0 : µ ≤ 230.7 versus Ha : µ > 230.7
e. H0 : µ ≤ 230 versus Ha : µ > 230
2. Find the value of the standardized test statistic.
a. 0.125
b. -0.125
c. 0.223
d. -0.7
e. 0.7
3. Find the P-value for the above mentioned test.
a. 0.8237
b. 0.0228
c. 0.5871
d. 0.4118
e. 0.0871

Answers

Answer 1
Answer:

Answer:

a) e) H0 : µ ≤ 230 versus Ha : µ > 230

b) t=(230.7-230)/((41.8)/(√(177)))=0.223    

c. 0.223

c) p_v =P(t_((176))>0.223)=0.4118  

d. 0.4118

Step-by-step explanation:

Information given  

\bar X=230.7 represent the sample mean

s=41.8 represent the sample standard deviation

n=177 sample size  

\mu_o =230 represent the value to verify

t would represent the statistic  

p_v represent the p value

Part a

We want to verify if the true mean is higher than 230 yards, the system of hypothesis would be:  

Null hypothesis:\mu \leq 230  

Alternative hypothesis:\mu > 230  

The best option would be:

H0 : µ ≤ 230 versus Ha : µ > 230

Part b

The statistic is given by:

t=(\bar X-\mu_o)/((s)/(√(n)))  (1)  

Replacing the info given we got:

t=(230.7-230)/((41.8)/(√(177)))=0.223    

Part c

The degrees of freedom are:

df=n-1=177-1=176  

The p value would be given by:

p_v =P(t_((176))>0.223)=0.4118  

d. 0.4118


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Answers

Answer:

x = 120

Step-by-step explanation:

An equilateral triangle has 3 equal angles

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

Vincent’s proportion is incorrect. His corresponding parts are not in the same position. The heights and bases are in different positions.

Step-by-step explanation:

I got it right on my assignment

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Step-by-step explanation: i got it right on my assignment

Maria found the least common multiple of 6 and 15. Her work is shown below.Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, . . .

Multiples of 15: 15, 30, 45, 60, . . .


The least common multiple is 60.


What is Maria’s error?

the answer is IDK

Answers

Answer:

The Least Common Multiple is 3

Step-by-step explanation:

Maria's error was that she tried finding the largest multiple rather than the least.

Answer:

Step-by-step explanation:

Maria's error on this is the LCM is 30 not 60

the least is 30 not 60

What is the volume of a cone with the given dimensions. radius=4 cm; height= 10 cm

Answers

167.552 the volume of a cone is piR^2h/3

Answer:

167.47 \:  {cm}^(3)

Step-by-step explanation:

V_(cone)  =  (1)/( 3) \pi {r}^(2)h \n  \n  =  (1)/( 3) \pi *  {4}^(2) * 10 \n  \n =  (1)/( 3)  * 3.14 * 16 * 10 \n  \n  =  (1)/( 3)  * \: 502.4 \n  \n  = 167.466667 \n  \n  = 167.47 \:  {cm}^(3)

The health of the bear population in yellowstone national park is monitored by periodic measurements. a sample of 54 bears has a mean weight of 182.9 lb. assuming the standard deviation is known to be 121.8lb find a 99% confidence intreval estimate

Answers

The 99% confidence interval will be found as follows:
x_bar +/- z(σ/√n)
where
z- z-score
σ-sigma
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thus from the information given we shall have:
182.9+/-2.58(121.8/√54)
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=225.6632 or 140.1368