Explain identifying like terms and unlike terms? how to solve combing like terms?

The pita‑franchise owner has observed in the past that waiting times tend to have a long tail to the right, with most customers

seropon [69]

Answer:

The two‑sample t ‑test is not appropriate because the distributions are not normal and the sample sizes are too small.

Step-by-step explanation:

The complete question is:

The owner of a pita franchise with two locations is interested in the average time that customers spend waiting for service at each store. She believes that the average waiting time at the original location is higher than the average waiting time at the new location.

The pita‑franchise owner has observed in the past that waiting times tend to have a long tail to the right, with most customers served relatively quickly and a few rare customers required to wait a very long time.Is a two‑sample t ‑test appropriate in this setting?The two‑sample t ‑test is appropriate because this is a comparison of the means of two continuous, random variables.The two‑sample t ‑test is not appropriate because the two samples do not have the same size.The two‑sample t ‑test is not appropriate because the sample standard deviations are not equal.The two‑sample t ‑test is not appropriate because the distributions are not normal and the sample sizes are too small.The two‑sample t ‑test is appropriate because the samples are random and contain no outliers, and the populations are normal.

For two sample t-test the distributions must be normal. Here the data, as mentioned in the questions is skewed to the right with long waiting times in the past.

7 0

4 years ago

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

Find the difference between 760 and 188

topjm [15]

Answer:

572

Step-by-step explanation:

Work :