1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Len [333]
2 years ago
10

The home run percentage is the number of home runs per 100 times at bat. A random sample of 43 professional baseball players gav

e the following data for home run percentages.
1.6 2.4 1.2 6.6 2.3 0 1.8 2.5 6.5 1.8
2.7 2 1.9 1.3 2.7 1.7 1.3 2.1 2.8 1.4
3.8 2.1 3.4 1.3 1.5 2.9 2.6 0 4.1 2.9
1.9 2.4 0 1.8 3.1 3.8 3.2 1.6 4.2 0
1.2 1.8 2.4

a. Use a calculator with mean and standard deviation keys to verify that x-bar = ~2.29 and s = ~1.40.
b. Compute a 90% confidence interval for the population mean mu of home run percentages for all professional baseball players.
c. Compute a 99% confidence interval for the population mean mu of home run percentages for all professional baseball players.
d. The home run percentages for three professional players are: Tim Huelett 2.5, Herb Hunter 2.0, and Jackie Jensen 3.8.
e. Examine your confidence intervals and describe how the home run percentages for these players compare to the population average.
f. In previous problems, we assumed the x distribution was normal or approximately normal. Do we need to make such an assumption in this problem? Why or why not?
Mathematics
1 answer:
Andru [333]2 years ago
6 0

Step-by-step explanation:

(a) Yes, if you enter all 43 values into your calculator, you calculator should report:

xbar = 2.293

s = 1.401

(b)

Note: Most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 1.684 * 1.401 / sqrt(43) = 1.933

2.293 - 1.684 * 1.401 / sqrt(43) = 2.653

Answer: (1.933, 2.653)

Note: To find the t-value that allows us to be 90% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.90)/2 = .05 or up from 90% depending on your t-table. So, the t-critical value is 1.684.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 1.681).

2.293 +/- 1.681 * 1.401 / sqrt(43)

(1.934, 2.652)

Note: Some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 90% CI is:

2.293 +/- 1.645 * 1.401 / sqrt(43)

(1.942, 2.644)

Note: To find the z-value that allows us to be 90% confident, (1) using the z-table, look up (1-.90)/2 = .05 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .05 or up from 90% depending on your t-table. Either way, the z-critical value is 1.645.

(c)

Note: Again, most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 2.704 * 1.401 / sqrt(43) = 1.715

2.293 - 2.704 * 1.401 / sqrt(43) = 2.871

Answer: (1.715, 2.871)

Note: To find the t-value that allows us to be 99% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.99)/2 = .005 or up from 99% depending on your t-table. So, the t-critical value is 2.704.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 2.698).

2.293 +/- 2.698 * 1.401 / sqrt(43)

(1.717, 2.869)

Note: Again, some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 99% CI is:

2.293 +/- 2.576 * 1.401 / sqrt(43)

(1.742, 2.843)

Note: To find the z-value that allows us to be 99% confident, (1) using the z-table, look up (1-.99)/2 = .005 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .005 or up from 99% depending on your t-table. Either way, the z-critical value is 2.576.

(d)

Tim Huelett 2.5

Since 2.5 falls between (1.715, 2.871), we see that Tim Huelett falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Herb Hunter 2.0

Since 2.0 falls between (1.715, 2.871), we see that Herb Hunter falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Jackie Jensen 3.8.

Since 3.8 falls above (1.715, 2.871), we see that Jackie Jensen falls in the 99% CI range. So, his home run percentage IS significantly GREATER than the population average.

(e)

Because of the Central Limit Theorem (CLT), since our sample size is large, we do NOT have to make the normality assumption since the CLT tells us that the sampling distribution of xbar will be approximatley normal even if the underlying population distribution is not.

You might be interested in
Please answer correctly !!!!!! Will mark brainliest !!!!!!!!!!!!!
Leni [432]

Answer:

At the time of launch height of the object was 60 meters.

Step-by-step explanation:

An object was launched from a platform and its height was modeled by the function,

h(x) = -5x² + 20x + 60

Where x = time or duration after the launch

At the time of launch, x = 0

So, by putting x = 0 in this equation,

h(0) = -5×(0) + 20×(0) + 60

h(0) = 60

Therefore, at the time of launch height of the object was 60 meters.

4 0
3 years ago
Graph f(x) = 2x^4+3 <br> let x = 0, 1, and 2.<br> please show your work
mestny [16]
Sub those points for x
f(x) is the y

(x,y)

so for x=0
f(0)=2(0)^4+3
f(0)=0+3
f(0)=3
(0,3) is a opint

for x=1
f(1)=2(1)^4+3
f(1)=2(1)+3
f(1)=2+3
f(1)=5
(1,5) is another point

for x=2
f(2)=2(2^4)+3
f(2)=2(16)+3
f(2)=32+3
f(2)=35
(2,35) is another point


points are
(0,3)
(1,5)
(2,35)
3 0
2 years ago
Need answer asap I’m timed!!!
Ugo [173]
Should be option 1. i cant really gv an explanation tho sry :/
3 0
3 years ago
A culture of bacteria obeys the law of uninhibited growth. if 500 bacteria are present initially, and there are 800 after 1 hour
lesantik [10]
Since this culture of bacteria obeys the law of uninhibited growth. Given that 500 bacteria are present initially, and there are 800 after 1 hour, a growth rate of 0.470003629246 (k value) has been calculated. <span>5243 </span>will be present after 5 hours.
8 0
3 years ago
Read 2 more answers
How do you solve the division problem 456÷5?
stiv31 [10]
To find the exact value using trigonometric identities
Exact form 456 divide a by 5
Decimal form 91.2
Mixed number form 91 1
-
5
4 0
2 years ago
Other questions:
  • Can someone please help me with this
    15·1 answer
  • A 4-billion-bushel corn crop brings a price of $2.4/bushel. A commodity broker uses the following rule of thumb: If the crop is
    9·1 answer
  • PLEASE HELP!! 15 POINTS
    13·1 answer
  • Use the equation Y = 3/27x-54 +5.
    10·2 answers
  • Sam is 10 years younger than one-half the age of his aunt. Let a represent his aunt’s age. Which shows his aunt’s age written as
    5·2 answers
  • PLEASE HELP
    13·2 answers
  • F(x) = -4x + 1; Find x = -1
    14·1 answer
  • Tony is going to watch a movie in his collection. He has 3 action movies, 11 comedies, and 9 dramas. He will randomly select one
    15·1 answer
  • Margaret has some goats. The goats produce an average total of 21.7 litres of milk per day for 280 days. Margaret sells the milk
    14·2 answers
  • Alan walks dogs on weekend she gets paid 8.50 per hour he said she works 5 hours and 45 minutes does she earn more or less than
    9·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!