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?

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.

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>

8

step-by-step explanation:

to find the amount of gift bags to use all the supplies, find the gcf or greatest common factor. this is the largest number which multiplies evenly into each term.

48 has factors: 1, 2, 3, 4, 6, 8, 12. 16, 24, 48

80 has factors: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80

64 has factors: 1, 2, 4, 8, 8, 16, 32, 64

the highest number factor they all share in common is 8. this means there will be 8 bags with 6 pencils, 10 markers, and 8 stickers in each bag.