Mathematics, 14.04.2020 17:37 tipbri6380

The mean caloric intake of an adult male is 2800 with a standard deviation of 115. To verify this information, a sample of 25 men are selected and determined to have a mean caloric intake of 2950. Determine the 98% confidence interval for mean caloric intake of an adult male. Solution: Since n < 30, a t-test must be used. 2950 – qt(1.98/2,24)115/sqrt(25) 2950 + qt(1.98/2,24)115/sqrt(25) [2892.68, 3007.32] What is wrong with this solution?

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Answer from: brooklyn4932

The confidence intervals express the distribution of sampling in a survey.

The 98% confidence interval is $\mathbf{(2892.684,3007.316)}$

The given parameters are:

$\mathbf{\mu = 2800}$ --- the population mean

$\mathbf{\bar x = 2950}$ --- the sample mean

$\mathbf{\sigma = 115}$ -- the standard deviation

$\mathbf{n = 25}$ --- the sample size

Calculate the degrees of freedom

$\mathbf{df = n - 1}$

$\mathbf{df = 25 - 1}$

$\mathbf{df = 24}$

Calculate the significance level

$\mathbf{\alpha /2=\frac{1 - 98\%}2 = 0.01}$

The critical value at $\mathbf{\alpha/2 = 0.01}$ and df = 24 is:

$\mathbf{t_{\alpha/2} = 2.492}$

Calculate the margin of error

$\mathbf{E = t_{\alpha/2} \frac{\sigma}{\sqrt n}}$

So, we have:

$\mathbf{E = 2.492 \times \frac{115}{\sqrt{25}}}$

$\mathbf{E = 2.492 \times \frac{115}{5}}$

$\mathbf{E = 2.492 \times 23}$

$\mathbf{E = 57.316}$

The confidence interval is then calculated as:

$\mathbf{CI =\bar x \pm E}$

So, we have:

$\mathbf{CI = 2950 \pm 57.316}}$

Split

$\mathbf{CI = (2950 - 57.316,2950 + 57.316)}$

$\mathbf{CI = (2892.684,3007.316)}$

So, the 98% confidence interval is $\mathbf{(2892.684,3007.316)}$

Read more about confidence intervals and margin of errors at:

link

Answer from: juliajordan427

Step-by-step explanation:

We would use the t- distribution.

From the information given,

Mean, μ = 2950

Standard deviation, σ = 115

number of sample, n = 25

Degree of freedom, (df) = 25 - 1 = 24

Alpha level,α = (1 - confidence level)/2

α = (1 - 0.98)/2 = 0.01

We will look at the t distribution table for values corresponding to (df) = 24 and α = 0.01

The corresponding z score is 2.492

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

2950 ± 2.492 × 115/√25

= 2950 ± 2.492 × 23

= 2950 ± 57.316

The lower end of the confidence interval is 2950 - 57.316 =2892.68

The upper end of the confidence interval is 2950 + 57.316 = 3007.32

The solution is correct.

Answer from: Quest

Benefiting gif hebr hen ben bb i don’t know if you want me or not but i’m just

Answer from: Quest

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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The mean caloric intake of an adult male is 2800 with a standard deviation of 115. To verify this in...