Paul recently purchased wii fit. each day wii fit calculates a wii fit age. over the course of the past week (n = 7 days), his wii fit age has a mean of 26.1 with a standard deviation of 6.2.
(a) calculate the 95% confidence interval for paul’s mean wii fit age.
(b) use the 95% confidence interval to determine whether paul’s age (35 years) is significantly different from his mean wii fit age and explain why. show your work.

a) [20.366, 31.834]

b) Since, The age of 35 years does not belongs to the calculated interval of 95% confidence interval (i.e [20.366, 31.834]).

Paul's age is significantly different from his Mean Wii Fit Age.

Step-by-step explanation:

Data provided:

Number of days i.e sample size, n = 7

Mean, μ = 26.1

Standard deviation, σ = 6.2

Degrees of freedom, df = n - 1 = 7 - 1 = 6

Confidence level = 95%,

Now,

For Confidence level of 95% , α = 1 - 0.95 = 0.05

= 0.025,

From the t stats table,

tc = t(α/2, df)

or

tc = t(0.025, 6) = 2.447

Therefore,

Margin of error =

or

Margin of error =

or

Margin of error = 5.734

Therefore,

The 95% confidence interval = Mean ± Margin of error

or

The 95% confidence interval = [ (26.1 - 5.734) , (26.1 + 5.734)]

or

The 95% confidence interval = [20.366, 31.834]

b) Since, The age of 35 years does not belongs to the calculated interval of 95% confidence interval (i.e [20.366, 31.834]).

Therefore, it can be concluded that the Paul's age is significantly different from his Mean Wii Fit Age.

answer: 10 turtles

step-by-step explanation: as turtles have four legs so 4.10=40 therefore birds are 4 which sue counted