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Leni [432]
3 years ago
10

A study of the use of social media asked a sample of 488 American adults under the age of 40 and a sample of 421 American adults

aged 40 or over about their use of social media. Based on their answers, each subject was assigned a social media usage score on a scale of 0 to 25. Higher scores indicate greater usage. The subjects were chosen by random digit dialing of telephone numbers. Are the conditions for two-sample t inference satisfied?
a. Maybe: the SRS condition is OK but we need to look at the data to check Normality
b. No: scores in a range between 0 and 25 cant be Normal
c. Yes: the STS condition is OK and large sample sized makes the Normality condition unnecessary.
Mathematics
1 answer:
mr_godi [17]3 years ago
8 0

Answer:

The correct option is (c).

Step-by-step explanation:

The assumptions of a two-sample <em>t</em>-test are:

  • Independent samples
  • Normality
  • Equal variances

It is provided that the sample of selected for American adults under the age of 40 is of size, <em>n</em>₁ = 488 and the sample of selected for American adults aged 40 or over is of size, <em>n</em>₂ = 421.

As the sample size is quite large the according to the Central limit theorem the sampling distribution of the sample mean would follow a Normal distribution.

And even if the sample was selected by dialing of various telephone numbers, the sample is not biased since the size of the sample is too large.

So, the sample can be considered as a simple random sample.

Thus, the conditions for two-sample t inference are satisfied.

The correct option is (c).

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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

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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

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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.

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