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?

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

The 98% confidence interval is

The given parameters are:

--- the population mean

--- the sample mean

-- the standard deviation

--- the sample size

Calculate the degrees of freedom

Calculate the significance level

The critical value at  and df = 24 is:

Calculate the margin of error

So, we have:

The confidence interval is then calculated as:

So, we have:

Split

So, the 98% confidence interval is

Read more about confidence intervals and margin of errors at:

link

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.

idk this answer sorry cant on this one.

step-by-step explanation: