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jek_recluse [69]
3 years ago
6

Someone please help

Mathematics
1 answer:
krok68 [10]3 years ago
4 0
I think the answer is B I hope I’m correct
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The home run percentage is the number of home runs per 100 times at bat. A random sample of 43 professional baseball players gav
Andru [333]

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.

6 0
3 years ago
When you<br> two integers with<br> sign,<br> you<br> and keep the
sweet [91]
Wait what sign do you mean
7 0
3 years ago
What is the volume of the following rectangular prism? 8 units 5 1/4 units2 ​
Pie

Answer:

volume equals l x w x h

8 x 5 1/4 x 2

42 x 2

84

4 0
3 years ago
Read 2 more answers
businessText message users receive or send an average of 62.7 text messages per day. How many text messages does a text message
KiRa [710]

Answer:

(a) The probability that a text message user receives or sends three messages per hour is 0.2180.

(b) The probability that a text message user receives or sends more than three messages per hour is 0.2667.

Step-by-step explanation:

Let <em>X</em> = number of text messages receive or send in an hour.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em>.

It is provided that users receive or send 62.7 text messages in 24 hours.

Then the average number of text messages received or sent in an hour is: \lambda=\frac{62.7}{24}= 2.6125.

The probability of a random variable can be computed using the formula:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0, 1, 2, 3, ...

(a)

Compute the probability that a text message user receives or sends three messages per hour as follows:

P(X=3)=\frac{e^{-2.6125}(2.6125)^{3}}{3!} =0.21798\approx0.2180

Thus, the probability that a text message user receives or sends three messages per hour is 0.2180.

(b)

Compute the probability that a text message user receives or sends more than three messages per hour as follows:

P (X > 3) = 1 - P (X ≤ 3)

              = 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

             =1-\frac{e^{-2.6125}(2.6125)^{0}}{0!}-\frac{e^{-2.6125}(2.6125)^{1}}{1!}-\frac{e^{-2.6125}(2.6125)^{2}}{2!}-\frac{e^{-2.6125}(2.6125)^{3}}{3!}\\=1-0.0734-0.1916-0.2503-0.2180\\=0.2667

Thus, the probability that a text message user receives or sends more than three messages per hour is 0.2667.

6 0
3 years ago
Could anyone help me with this
Romashka-Z-Leto [24]
The answer is 154

This is because you add all the sides

33+56+65=154
8 0
4 years ago
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