Question

Educational Television: In a random sample of 200 people, 154 said that they watched educational television. Find the 90% confidence interval of the true proportion of people who watched educational television. Then write in a z interval

Answer #1

In a random sample of 200 people, 154 said that they watched
educational television. Find the 90% confidence interval of the
true proportion of people who watched educational television. If
the television company wanted to publicize the proportion of
viewers, do you think it should use the 90% confidence
interval?

A random sample of 803 people was taken. 154 of the people in
the sample favored Candidate A. Find 90% confidence interval for
the true proportion of people who favors Candidate A. Enter in the
lower limit of the confidence interval you found.

A random sample of 994 people was taken. 262 of the people in
the sample favored Candidate A. Find 90% confidence interval for
the true proportion of people who favors Candidate A. Enter in the
lower limit of the confidence interval you found.

In a recent poll, 600 people were asked if they liked soccer,
and 10% said they did. Based on this, construct a 90% confidence
interval for the true population proportion of people who like
soccer.
As in the reading, in your calculations:
--Use z = 1.645 for a 90% confidence interval
--Use z = 2 for a 95% confidence interval
--Use z = 2.576 for a 99% confidence interval.
Give your answers as decimals, to 4 decimal places.
[________, ________]

In a recent poll, 300 people were asked if they liked soccer,
and 37% said they did. Based on this, construct a 90% confidence
interval for the true population proportion of people who like
soccer.
As in the reading, in your calculations:
--Use z = 1.645 for a 90% confidence interval
--Use z = 2 for a 95% confidence interval
--Use z = 2.576 for a 99% confidence interval.
Give your answers as decimals, to 4 decimal places.
[,]

In a random sample of 250 Winnipeggers, 80 of them said they use
transit. A 94% confidence interval for the true proportion of all
Winnipeggers who use transit is:
Note: Value of 90%(1.645 and 95%(1.960) is only provided in the
table.

1. In a random sample of 200 college graduated, 42 said they
think a college degree is not worth the cost. Test to see if this
sample provides significant evidence that the population proportion
of college graduated who believe a college degree is not worth the
cost is different from 25%. Use a 5
a.
b. Calculate the observed sample statistic:
Calculate the z-scores for this sample statistic: z=
________
c. Use Statkey to find the p-value for this test:...

1) A media analyst is trying to determine the proportion of
people living in St. Louis that watch a particular television show.
The analyst collects a random sample of 204 people from St. Louis.
Of the 204 people in the sample, 56 people said that they did watch
the show.
a) Find a 95% confidence interval for the true population
proportion of people in St. Louis who watch the show.
b) Provide the right endpoint of the interval
as your...

In an simple random sample of 200 college students, 114 said
that they usually speed on interstate highways by 10 mph or more.
Give a 95% confidence interval for the corresponding proportion for
all college students. a: (0.5014,0.6455) b: (0.5431,0.6386) c:
(0.3567,0.4533) d: (0.5014,0.6386)

In the run-up to the 2016 election, a random sample of 100
people was taken to determine a confidence interval for the
population proportion who would vote in favor of Proposition 61 to
legalize marijuana. The confidence interval was found to
be (.402 to .598).
What was the confidence level, (1-alpha)*100%, associated with
this confidence interval?
95%
90%
We would need to know the number in the sample who said they
would vote for Proposition 61. If we knew this number,
we...

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