Help me on Stat problem please!?

Suppose a survey revealed that 20% of 491 respondents said they had in the past sold unwanted gifts over the Internet.

(a) Use the information to construct a 90% confidence interval for the population proportion who sold unwanted gifts over the Internet, rounding your margin of error to the nearest hundredth. (Round your answers to two decimal places.)

( , )

(b) Use the information to construct a 98% confidence interval for the population proportion who sold unwanted gifts over the Internet, rounding your margin of error to the nearest hundredth. (Round your answers to two decimal places.)

( , )

(c) Which interval has a higher probability of containing the true population proportion?

the 90% confidence interval

the 98% confidence interval

(d) Which interval is narrower?

the 90% confidence interval

the 98% confidence interval

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  • 8 years ago
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    A) These are proportion problems. You've been given the sample proportion information. We need to find the following information for a confidence interval:

    CI = Point Estimate +/- Critical Value ( Standard Error)

    Our Point Estimate is our Sample mean = .2

    Our Critical Value is a Z-score that will represent the low and high bound of 5% on each side (keep in mind, Confidence being 90% means we have 10% left over. That is split over the right and left side). To find this score, we need to use invnorm on the TI-83/84. Goto 2nd/vars and choose invnorm. Input the following: invnorm(.95). This is your critical value. We choose .95 because of the 10% left over from our CI we have to split that in half. We add HALF of the leftover % into the invnorm calculation. We will get z = 1.645

    Now, we need to find the standard error. That formula is as follows:

    SQRT ((p(1-p)/n))

    p = sample mean

    n = sample size

    Plug in the #'s and solve. You get .018

    Now, let's plug in our info to our formula from above.

    CI = .2 +/- 1.645(.018) = (.171, .229)

    This means, we are 90% confident that the interval of (.171, .229) contains the true proportion of respondents who claim they sold unwanted gifts in the past.

    B) 98% CI

    CI = Point Estimate +/- Critical Value ( Standard Error)

    Our Point Estimate is our Sample mean = .2

    Our Critical Value is a Z-score. Recall that we're using 98% CI...meaning we have 1% left on both sides. We use invnorm(.99) to find z = 2.33

    Now, we need to find the standard error. That formula is as follows:

    SQRT ((p(1-p)/n))

    p = sample mean

    n = sample size

    Plug in the #'s and solve. You get .018

    Now, let's plug in our info to our formula from above.

    CI = .2 +/- 2.33(.018) = (.158, .242)

    This means, we are 98% confident that the interval of (.158, .242) contains the true proportion of respondents who claim they sold unwanted gifts in the past.

    (c) Which interval has a higher probability of containing the true population proportion?

    the 98% confidence interval - The higher the level of confidence, the more likely the interval will contain the true population proportion.

    (d) Which interval is narrower?

    the 90% confidence interval - 90% is smaller confidence than 98%. Therefore the interval is more narrow.

    Source(s): AP Stats teacher.
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