Assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) find the critical value t
Subscript alpha divided by 2, (b) find the critical value z Subscript alpha divided by 2, or (c) state that neither the normal distribution nor the t distribution applies.
The confidence level is 95%, sigma is not known, and the histogram of 60 player salaries (in thousands of dollars) of football players on a team is as shown.
The confidence level is 95%, sigma is not known, and the histogram of 60 player salaries (in thousands of dollars) of football players on a team is as shown.
1 answer:
Answer:
Neither the normal distribution nor the t distribution applies.
Step-by-step explanation:
Hello!
The histogram attached shows the salary (in dollars) of 63 football players.
As you can see most of the values of this distribution are around 0 and the distribution is extremely asymmetrical (right skewed).
To apply a t-test to study the population average the population has to have a normal distribution.
The same goes for the standard normal distribution, but in this case, considering that the sample is large enough (n≥30) you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal X[bar]≈N(μ;δ²/n), but since the population variance is unknown, this is also not applicable.
The correct choice is the last one, since the conditions to apply the t test and the standard normal are not met.
I hope this helps!
You might be interested in
Answer:
D
Step-by-step explanation:
The augmented matrix for the system of three equaitons is
Multiply the first row by 5, the second row by -3 and add these two rows:
Subtract the third row from the second:
Divide the third row by 6:
Now multiply the third equation by 26 and add it to the second row:
You get the system of three equations:
From the third equation
Substitute z=2 into the second equation:
Now substitute z=2 and y=5 into the first equation:
The solution is (1,5,2)
For each <em>x</em> in the interval 0 ≤ <em>x</em> ≤ 5, the shell at that point has
• radius = 5 - <em>x</em>, which is the distance from <em>x</em> to <em>x</em> = 5
• height = <em>x</em> ² + 2
• thickness = d<em>x</em>
and hence contributes a volume of 2<em>π</em> (5 - <em>x</em>) (<em>x</em> ² + 2) d<em>x</em>.
Taking infinitely many of these shells and summing their volumes (i.e. integrating) gives the volume of the region:
