A group of GSU students in the Young Democrats Club wish to determine the likeability of their favorite candidate. They survey 2322 randomly

Question

A group of GSU students in the Young Democrats Club wish to determine the likeability of their favorite candidate. They survey 2322 randomly selected registered voters and ask them to rate their candidate (on a “thermometer” from 0 to 100, where 0 means “very cold”and 100 means “very warm”feelings).Suppose the sample of n=2322 responses have a mean warmth of 64.16 with a sample standard deviation of s=26.34.Find the 95% confidence interval to estimate the population/national mean warmth rating for their candidate.A) Note: The critical value, tc, that we use to calculate the margin of error for a 95% Confidence Interval is tc = ____ (round to 4 decimal places)B) The margin of error for this confidence interval is: m = ____ (round to 4 decimal places)C) The 95% Confidence Interval for the likeability of their candidate is: (____ , ____) round each to 2 decimal places

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Ivy 2 days 2021-10-10T21:27:19+00:00 1 Answer 0

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    2021-10-10T21:29:17+00:00

    Answer:

    a) tc=\pm 1.9601

    b) m=1.9601 \frac{26.34}{\sqrt{2322}}=1.0714

    c) The 95% confidence interval is given by (63.09;65.23)

    Step-by-step explanation:

    1) Notation and definitions

    n=2322 represent the sample size

    \bar X= 64.16 represent the sample mean

    s=26.34 represent the sample standard deviation

    m represent the margin of error

    A confidence interval is “a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval”.  

    The margin of error is the range of values below and above the sample statistic in a confidence interval.  

    Normal distribution, is a “probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean”.  

    2) Calculate the critical value tc

    In order to find the critical value is impornta to mention that we don’t know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. The degrees of freedom are given by:

    df=n-1=2322-1=2321

    We can find the critical values in excel using the following formulas:

    “=T.INV(0.025,2321)” for t_{\alpha/2}=-1.9601

    “=T.INV(1-0.025,2321)” for t_{1-\alpha/2}=1.9601

    The critical value tc=\pm 1.9601

    3) Calculate the margin of error (m)

    The margin of error for the sample mean is given by this formula:

    m=t_c \frac{s}{\sqrt{n}}

    m=1.9601 \frac{26.34}{\sqrt{2322}}=1.0714

    4) Calculate the confidence interval

    The interval for the mean is given by this formula:

    \bar X \pm t_{c} \frac{s}{\sqrt{n}}

    And calculating the limits we got:

    64.16 - 1.9601 \frac{26.34}{\sqrt{2322}}=63.09

    64.16 + 1.9601 \frac{26.34}{\sqrt{2322}}=65.23

    The 95% confidence interval is given by (63.09;65.23)

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