What fractional part of 1 pound is an ounce?

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streamin

oksian1 [2.3K]

Answer:

The 99% confidence level for the proportion of all adult Americans who watched streamed programming up to that point in time is (0.514, 0.566). This means that we are 99% sure that the true proportion of all American adults surveyed said they have watched digitally streamed TV programming on some type of device is between 0.514 and 0.566.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

A poll reported that 54% of 2342 American adults surveyed said they have watched digitally streamed TV programming on some type of device.

This means that $\pi = 0.54, n = 2342$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.54 - 2.575\sqrt{\frac{0.54*0.46}{2342}} = 0.514$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.54 + 2.575\sqrt{\frac{0.54*0.46}{2342}} = 0.566$

The 99% confidence level for the proportion of all adult Americans who watched streamed programming up to that point in time is (0.514, 0.566). This means that we are 99% sure that the true proportion of all American adults surveyed said they have watched digitally streamed TV programming on some type of device is between 0.514 and 0.566.

6 0

3 years ago

If there are 4 hunting 7 fishing 6 hiking 3 camping how many outdoor expedition are possible

finlep [7]

Answer:

20

Step-by-step explanation:

6 0

3 years ago

An experiment was performed to compare the abrasive wear of two different laminated materials. Twelve pieces of material 1 were

vladimir1956 [14]

Answer:

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

There is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units

Step-by-step explanation:

Step:-(1)

Given data the samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4

Mean of the first sample x₁⁻ =85

standard deviation of the first sample S₁ = 4

Given data the samples of material 2 gave an average of 81 and a sample standard deviation of 5.

Mean of the first sample x₂⁻ =81

standard deviation of the first sample S₂ = 5

Step :-2

Null hypothesis: H₀: there is no significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Alternative hypothesis :H₁: there is significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Assume the populations to be approximately normal with equal variances.σ₁² =σ₂²

The test statistic

$Z= \frac{x_{1} -x_{2} }{\sqrt{\frac{S^2_{1} }{n_{1} } +\frac{S^2_{2} }{n_{2} } } }$

Given n₁=n₂=60.

$Z= \frac{85-81 }{\sqrt{\frac{(4)^2 }{60 } +\frac{5^2 }{60} } }$

On calculation, we get

Z = $\frac{4}{\sqrt{0.6833} }$

z = 4.8389

The tabulated value Z =1.96 at 0.05 level of significance.

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

Conclusion:-

there is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units.

3 0

3 years ago

Carl and Cameron are at a pool. Carl dives in and touches the bottom of the pool, 15 feet below the surface of the water.

Sergio039 [100]

Answer:

17 feet apart

Step-by-step explanation:

4 0

3 years ago

Martin tosses 4 pennies. What is the probability that at least one of the coins lands on heads? At least 3 coins land on heads?

EleoNora [17]

Answer: 15/16

there are 15 scenarios in which you get at least 1 head and one where you don't therefore the answer is 15/16

6 0

3 years ago