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nordsb [41]
2 years ago
12

In a survey, 18 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped

with a sample mean of $47 and sample standard deviation of $3. Construct a confidence interval at a 90% confidence level.
Mathematics
1 answer:
AleksAgata [21]2 years ago
5 0

Answer:

The 90% confidence interval for the mean amount spent on the children's birthday gift is between $45.77 and $48.23.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 18 - 1 = 17

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 17 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.9}{2} = 0.95. So we have T = 1.7396

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.7396\frac{3}{\sqrt{18}} = 1.23

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 47 - 1.23 = $45.77

The upper end of the interval is the sample mean added to M. So it is 47 + 1.23 = $48.23

The 90% confidence interval for the mean amount spent on the children's birthday gift is between $45.77 and $48.23.

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erik [133]

Answer:

set mary is x years old, and helen is y years old.

first sentence:x=2y

second one:4(y-8)=x+4

bring x=2y in 4(y-8)=x+4

=>4y-32=2y+4

=>2y=36

=>y=18

=>x=36

mary:36

helen:18

3 0
3 years ago
(a) determine the null and alternative hypotheses, (b) explain what it would mean to make a Type I error, and (c) explain what i
iren2701 [21]
Null: the mean amount of peanut butter in a jar is equal to 32 oz.
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type 2 error is failing to reject the null when it is actually false. this means that we would say the mean amount of peanut butter is equal to 32 when in reality it is less.
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3 years ago
A textbook search committee is considering
lora16 [44]

Number of ways to select 3 books from 13 books for adoption is 286 .

<u>Step-by-step explanation:</u>

A Permutation is an ordered Combination. When the order does matter it is a Permutation. There are basically two types of permutation:

  • Repetition is Allowed: such as the lock above. It could be "333".
  • No Repetition: for example the first three people in a running race. You can't be first and second.

Formula is given by:

nC_r= \frac{n!}{r! (n-r)!} , where n is the number of things to choose from,  and we choose r of them,  no repetitions,  order matters. Here , n=13 , r=3.

⇒ 13C_3 = \frac{13!}{3!(13-3)!}

⇒ 13C_3 = \frac{13!}{3!(10)!}

⇒ 13C_3 = \frac{13(12)(11)10!}{3(2)(1)(10)!}

⇒ 13C_3 = 13(2)(11)

⇒ 13 C_3 = 286

∴ Number of ways to select 3 books from 13 books for adoption is 286 .

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12 and 15 

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