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Zanzabum
3 years ago
8

The life in hours of a battery is known to be approximately normally distributed, with standard deviation σ = 1.25 hours. A rand

om sample of 10 batteries has a mean life of x overBar equals 40.5 hours.
(a) Is there evidence to support the claim that battery life exceeds 40 hours? Use α = 0.020.
(b) What is the P-value for the test in part (a)?
(c) What is the β-error for the test in part (a) if the true mean life is 42 hours?
(d) What sample size would be required to ensure that β does not exceed 0.1 if the true mean life is 44 hours?
Round your answers to three significant digits.


(a) The battery life ( IS ??? or IS NOT ??? ) significantly different greater than 40 hours at α = 0.020.
(b)p-value = _____?_____ Round your answer to three decimal places (e.g. 98.765).
(c)β = _____?_____ Round your answer to five decimal places (e.g. 98.76543).
(d) ____?_____batteries

Mathematics
1 answer:
Papessa [141]3 years ago
6 0

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

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Answer:

1) Is the pattern an arithmetic sequence?​

Yes it is

2)Identify a and d.​

a = First term = 2

d = Common difference = 2

3) Write the 50th term of the sequence.​

50th term = 100

4) Find the total number of logs in the first 10 rows.​

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Step-by-step explanation:

Is the pattern an arithmetic sequence?​

Yes it is

2) Identify a and d.​

A pyramid of logs has 2 logs in the top row, 4 logs in the second row from the top, 6 logs in the third row from the top, and so on,

The formula for arithmetic sequence =

an = a+ (n - 1)d

a = First term

d = Common difference

For the above question:

a = 2

d = Second term - First term

= 4 - 2

d = 2

3) Write the 50th term of the sequence.​

Using the formula for arithmetic sequence

an = a+ (n - 1)d

a = 2

n = 50

d = 2

a50 = 2 + (50 - 1)2

= 2 + (49)2

= 2 + 98

= 100

The 50th term = 100

4)Find the total number of logs in the first 10 rows.​

Sum of first n terms = n/2(a + l)

n = 10

a = first term = 2

We are told that there are 200 logs in the bottom row, hence:

l = last term = 200 logs

Hence,

Sn = 10/2×[ (2 + 200

= 5(202)

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