Every square meter of ceiling needs 10.75 tiles. To calculate the number of tiles needed for various ceiling areas: For a ceiling area of 1, it requires 10.75 tiles. For a ceiling area of 10, it requires 10.75 × 10, which equals 107.5 tiles. For an area of 100, it needs 10.75 × 100, totaling 1075 tiles. Consequently, that represents the necessary solution.

6 0

1 month ago

Question 1 Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your metho

babunello [11817]

Answer:

(a) The 95% confidence interval representing the percentage of the entire U.S. population that would select American football as their preferred television sport is (0.34, 0.40).

(b) Not reasonable.

Step-by-step explanation:

Given:

n = 1000

$\hat p$ = 0.37

(a)

The confidence interval (1 - α)% for proportion p is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

Where:

$\hat p$ = the sample proportion

n = sample size

$z_{\alpha/2}$ = critical z value.

Calculate the critical value of z for a 95% confidence level as shown:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Refer to a z-table for the necessary value.

Compute the 95% confidence interval for proportion p as follows:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$=0.37\pm 1.96\times\sqrt{\frac{0.37(1-0.37)}{1000}}$

$=0.37\pm 0.03\\=(0.34, 0.40)$

Thus, the 95% confidence interval indicating the proportion of individuals in the U.S. who might say their preferred sport on TV is American football is (0.34, 0.40).

(b)

Next, we must assess if it's rational to consider that the actual percent of people in the United States who favor American football on television is 33%.

The hypothesis is defined as:

H₀: The portion of the U.S. population claiming American football as their favorite sport on television is 33%, meaning p = 0.33.

Hₐ: The proportion of people in the U.S. whose favorite sport to watch on television is not 33%, or p ≠ 0.33

This hypothesis can be verified using a confidence interval.

The decision rule:

If the (1 - α)% confidence interval contains the null value of the hypothesis, then the null hypothesis is not rejected. If, however, the (1 - α)% confidence interval excludes the null value of the hypothesis, then the null hypothesis is rejected.

The confidence interval does encompass the null value of p, which is 0.33.

<pthus the="" null="" hypothesis="" will="" be="" rejected.=""><pin conclusion="" it="" is="">not reasonable to accept that 33% represents the actual percentage of those in the U.S. whose favorite televised sport is American football.

</pin></pthus></pthe>

6 0

3 months ago

Let u = <5, 6>, v = <-2, -6>. Find -2u + 5v.

babunello [11817]

-2(5,6)-40 That should cover it.

7 0

2 months ago

According to government​ data, the probability that an adult was never in a museum is 10​%. In a random survey of 60 ​adults, wh

According to government data, the probability that an adult was never in a museum is 10%. In a random survey of 60 adults, wh