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Alexxx [7]
2 years ago
8

A psychologist takes an SRS of 404040 people in their twenties and 606060 people in their fifties and surveys them about their a

nxieties. Suppose that 48\%48%48, percent of people in their twenties and 28\%28%28, percent of people in their fifties have anxiety about flying. The psychologist will then look at the difference (\text{20's}-\text{50's})(20’s−50’s)left parenthesis, start text, 20, apostrophe, s, end text, minus, start text, 50, apostrophe, s, end text, right parenthesis between the proportions of people with anxiety about flying in each sample. What will be the shape of the sampling distribution of the difference in sample proportions, and why?
Mathematics
1 answer:
Leto [7]2 years ago
3 0

Answer: Approximately normal, because we expect 19.2 successes and 20.8 failures from people in their twenties, and 16.8 and 43.2 from people in their fifties, and all of these counts are at least 10.

Step-by-step explanation:

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Answer:

a) For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

t_{\alpha/2} =\pm 2.35

b) For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

t_{\alpha/2} =\pm 2.62

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.  

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

Part a

For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

t_{\alpha/2} =\pm 2.35

Part b

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

t_{\alpha/2} =\pm 2.62

3 0
3 years ago
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