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notka56 [123]
3 years ago
9

Suzanne owns a small business that employs 555 other people. Suzanne makes \$100{,}000$100,000dollar sign, 100, comma, 000 per y

ear, and the other 555 employees make between \$40{,}000$40,000dollar sign, 40, comma, 000 and \$50{,}000$50,000dollar sign, 50, comma, 000 per year.
Suzanne decides to increase her salary by \$30{,}000$30,000dollar sign, 30, comma, 000 per year and leave the rest of the salaries the same.
How will increasing her salary affect the mean and median? How will increasing her salary affect the mean and median?
Mathematics
2 answers:
zalisa [80]3 years ago
6 0

Answer:

16

Step-by-step explanation:

Tpy6a [65]3 years ago
6 0

Answer:

Mean increased

Median remains same

Step-by-step explanation:

Suzanne make $ 100 k per year

5 other employee  $ 40 k  to $ 50 K per year per employee

data in ascending order

$40000  , $40000+a  , $40000+b  , $40000+c   , $50000  , $ 100000

where 10000 ≥c ≥ b ≥ a ≥ 0

Mean = ($40000 + $40000+a + $40000+b  + $40000+c  + $50000 + $ 100000) /6

Mean = $(310000 + a + b + c)/ 6

Median = ($40000+b  + $40000+c )/2 =  $40000 + (b+c)/2

Suzanne make now $ 100000 + $  30000 = $ 130000 per year

$40000  , $40000+a  , $40000+b  , $40000+c   , $50000 , $ 130000

Mean = ($40000 + $40000+a + $40000+b  + $40000+c  + $50000 + $ 130000) /6

Mean = $(340000 + a + b + c)/ 6

Median = ($40+b k + $40+c k)/2 =  $40 + (b+c)/2 k

Mean increased by  $(340000 + a + b + c)/ 6 - $(310000 + a + b + c)/ 6

= $ 30000/6  = $ 5000

Median Remains the same

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A student carried out an experiment to determine the amount of vitamin C in a tablet sample. He performed 5 trials to produce th
ivolga24 [154]

Answer:

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

P-value = 0.166.

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{5}(490+502+505+495+492)\\\\\\M=\dfrac{2484}{5}\\\\\\M=496.8\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{4}((490-496.8)^2+(502-496.8)^2+(505-496.8)^2+(495-496.8)^2+(492-496.8)^2)}\\\\\\s=\sqrt{\dfrac{166.8}{4}}\\\\\\s=\sqrt{41.7}=6.5\\\\\\

Then, we can perform the hypothesis t-test for the mean.

The claim is that the amount of vitamin C in a tablet sample is different from 500 mg.

Then, the null and alternative hypothesis are:

H_0: \mu=500\\\\H_a:\mu< 500

The significance level is 0.05.

The sample has a size n=5.

The sample mean is M=496.8.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.5.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{6.5}{\sqrt{5}}=2.907

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{496.8-500}{2.907}=\dfrac{-3.2}{2.907}=-1.1

The degrees of freedom for this sample size are:

df=n-1=5-1=4

This test is a left-tailed test, with 4 degrees of freedom and t=-1.1, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t

As the P-value (0.166) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

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