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
larisa86 [58]
2 years ago
9

A sample of 18 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is n

ormal. The weight of each bag was then recorded. The mean weight was 3 ounces with a standard deviation of 0.15 ounces. The population standard deviation is known to be 0.1 ounce.
A.
1. bar x =____.
2. sx =____.
B. In words, define the random variable .
C. In words, define the random variable .
D. Which distribution should you use for this problem? Explain your choice.
E. Construct a 90 confidence interval for the population mean weight of the candies.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
F. Construct a 98 confidence interval for the population mean weight of the candies.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
G. In complete sentences, explain why the confidence interval in part f is larger than the confidence interval in part e.
H. In complete sentences, give an interpretation of what the interval in part f means.
Mathematics
1 answer:
ki77a [65]2 years ago
5 0

Answer:

a)

i.  x=- - - - - -

ii.  o=- - - - - -

iii.  Sx=- - - - -

b) The random variable X signifies the weights of the bags of candies which are selected at random.  

c) The statistics variable X  is a measure on a sample that is used as an estimate of the population mean.X  is the mean weight of the 16 bags of candies that were selected.

d) The distribution needed for this problem is normal distribution with parameters  because the population standard deviation is known.

N(X, o/\sqrt{n})

So, the distribution is  

N(2, 0.12/\sqrt{16})

e)

i. The 90% confidence interval for the population mean weight of the candies is  1.9589 , 2.0411

iii. The error bound is 0.0411

f)

i. The 98% confidence interval for the population mean weight of the candies is  1.9418 , 2.0582

iii. The error bound is 0.0582

g) The difference in the confidence intervals in part (f) and part (e) is of the level of confidence. The change is, the change in the area being calculated for the normal distribution. Therefore, the larger confidence level results in larger area and larger interval.

Hence the interval in part  is larger than the one in part (e).

h) The interval in part (f) signifies that with 98% confidence that the population mean weight of the bag of candies lies between 1.942 ounce and 2.058 ounce.

Step-by-step explanation:

a)

i.  x=- - - - - -

ii.  o=- - - - - -

iii.  Sx=- - - - -

b) The random variable X signifies the weights of the bags of candies which are selected at random.  

c) The statistics variable X  is a measure on a sample that is used as an estimate of the population mean.X  is the mean weight of the 16 bags of candies that were selected.

d) The distribution needed for this problem is normal distribution with parameters  because the population standard deviation is known.

N(X, o/\sqrt{n})

So, the distribution is  

N(2, 0.12/\sqrt{16})

e)

i. The 90% confidence interval for the population mean weight of the candies is  1.9589 , 2.0411

iii. The error bound is 0.0411

f)

i. The 98% confidence interval for the population mean weight of the candies is  1.9418 , 2.0582

iii. The error bound is 0.0582

g) The difference in the confidence intervals in part (f) and part (e) is of the level of confidence. The change is, the change in the area being calculated for the normal distribution. Therefore, the larger confidence level results in larger area and larger interval.

Hence the interval in part  is larger than the one in part (e).

h) The interval in part (f) signifies that with 98% confidence that the population mean weight of the bag of candies lies between 1.942 ounce and 2.058 ounce.

You might be interested in
Choose where it is a function or not a function also choose what is the range and the domain
kow [346]
Not a function because of the vertical line test.
Domain is [-2,∞] since the left most point is -2 and the right is unbounded.
Range is (-∞,∞) since it is not bounded in terms of y.
7 0
3 years ago
the names of nine boys and two girls are in a hat. what is the probability of first drawing a girls name and then a boys name fr
lara [203]
18/121 shoul be the answer
5 0
2 years ago
Read 2 more answers
16 multiply 7/8 in simplest
denpristay [2]
Answer: 14

= 16 x (7/8)
= 8 goes into 16 2 times
= 2 x 7
= 14
5 0
2 years ago
Problem page the ratio of men to women working for a company is 5 to 8 . if there are 169 employees total, how many women work f
Ksenya-84 [330]
5 : 8....added = 13

5/13(169) = 845/13 = 65 men
8/13(169) = 1352/13 = 104 women <===
8 0
3 years ago
A veterinarian wants to know if pit bulls or golden retrievers have a higher incidence of tooth decay at the age of three. The v
BigorU [14]

Answer: B) [-0.0332,0.1332]

Step-by-step explanation: <u>Confidence</u> <u>Interval</u> is an interval where we can be a percentage sure the true mean is.

The confidence interval for a difference in population proportion is calculated following these steps:

First, let's find population proportion for each population:

p_{1}=\frac{30}{120}=0.25

p_{2}=\frac{32}{160}=0.2

Second, calculate standard deviation for each proportion:

\sigma_{1}=\sqrt{\frac{0.25(0.75)}{120} } = 0.0395

\sigma_{2}=\sqrt{\frac{0.2(0.8)}{160} } = 0.0316

Now, we calculate standard error for difference:

SE=\sqrt{\sigma_{1}^{2}+\sigma_{2}^{2}}

SE=\sqrt{0.0395^{2}+0.0316^{2}}

SE = 0.0505

The z-score for a 90% CI is 1.645.

Then, confidence interval is

p_{1}-p_{2} ± z-score.SE

0.25-0.2 ± 1.645(0.0505)

0.05 ± 0.0831

The limits of this interval are:

inferior: 0.05 - 0.0831 = -0.0332

superior: 0.05 + 0.0831 = 0.1332

The 90% confidence interval for the difference in the population proportion of pit pulls and golden retrievers is [-0.0332,0.1332].

8 0
3 years ago
Other questions:
  • X2 – 5x + 6 = 0 solve.
    9·1 answer
  • Umesh’s total charge for an onsite job is: a fixed charge of $50 per job, plus an hourly charge of $80 per hour. Which one of th
    10·1 answer
  • -x-9 if x &lt;-2<br> 23. h(x) = (x +1)-5 if -25x&lt;2<br> 12x - 3) if x 2
    11·1 answer
  • How do you solve 4a/b divided by 2ac/b
    13·2 answers
  • Can 2yards and 60 inches be expressed as a ratio if so express the ratio in simplest form
    6·1 answer
  • Find the LCM for 6 and 4.
    14·2 answers
  • If b=a-c are supplementary angle find the value of a​
    5·1 answer
  • Someone help me solve these two please !!
    13·2 answers
  • Give the linear inequality shown in the graph below.
    6·1 answer
  • Find the total amount you will pay given an original price of $499.95 and tax rate of 4%
    7·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!