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Keith_Richards [23]
3 years ago
15

Use the formula to find the standard error of the distribution of differences in sample means, x¯1-x¯2 . Samples of size 110 fro

m Population 1 with mean 94 and standard deviation 14 and samples of size 85 from Population 2 with mean 72 and standard deviation 15 Round your answer for the standard error to
Mathematics
1 answer:
Volgvan3 years ago
7 0

Answer:

2.104

Step-by-step explanation:

Data provided in the question:

For sample 1,

n₁ = 110

μ₁ = 94

s₁ = 14

For sample 2,

n₂ = 85

μ₂ = 72

s₂ = 15

Now,

Standard error = \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}

or

Standard error = \sqrt{\frac{14^2}{110}+\frac{15^2}{85}}

or

Standard error = √4.43

or

Standard error = 2.104

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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
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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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What is the value of yearly compensation package that includes $65,000 salary the forecast of a $350 Dollar per month health ins
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Answer:

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Step-by-step explanation:

The yearly compensation package includes a yearly salary of $65000, the health insurance plan of $350 per month and a life insurance premium of $30 per month.

Total compensation in health insurance plan is $(350 × 12) = $4200

Total compensation in life insurance premium is $(30 × 12) = $360

Therefore, total yearly compensation package is $(65000 + 4200 + 360) = $69560 (Answer)

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