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Thread: Statistics help

  1. #1
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    Dec 2009

    Statistics help

    A researcher asks whether attending a private high school leads to higher or lower performance on a test of social skills when compared to students attending public schools. A sample of 100 students from a private school produces a mean score of 71.30. The population mean (m) for students from public high schools is 75.62. The population standard deviation is 28. Zobt is 1.54. Zcrit is 1.96.
    Should the researcher use a one-tailed or a two-tailed test? Why?
    What is the alternative hypothesis?
    What is the null hypothesis?
    What should the researcher conclude about this relationship in the population?
    Are the results significant? Explain your response.
    What is the probability of making a Type I error?
    If a Type I error were made, what would it mean?
    What is the probability of making a Type II error?
    If a Type II error were made, what would it mean?

  2. #2
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    Nov 2009

    Re: Statistics help

    (1) The researcher should not use a two-tailed test if wants to differentiate between higher and lower results,
    as this simply tests whether or not the results are significantly different.
    If "higher or lower" is referring to the fact that the researcher is not concerned whether the results
    are specifically higher or specifically lower, but fall in either bracket,
    then a two-tailed test suffices.
    If the researcher wants to know if the results from the private school are down
    compared to the results from the public schools, he should check for the mean result
    of the private school students being significantly less than the mean result of the public school students.
    A one-tailed test is used for this.
    (If he wants to know if the results are up, which the data does not point to,
    he should check for the private school mean being greater than the public school mean.)

    The way the question is worded "suggests" a two-tailed test if the researcher is not examining the directional bias.

    (2) The alternative hypothesis, H[sub9367syt]a[/sub9367syt] is....
    this sample data from the private school does not appear to be the same
    as a sample expected from the overall population of the public schools.

    This means that the alternative hypothesis states that the sample results from the private school
    suggests a difference between the schools in terms of performance.

    (3) The null hypothesis, H[sub9367syt]0[/sub9367syt] is....
    the sample appears the same as one that belongs to the public schools population,
    meaning that there is no significant difference between the public school results and the private school results.

    (4) Z[sub9367syt]obt[/sub9367syt] has been calculated from the data.
    This z-value needs to be beyond the critical values to be considered as "not belonging" to the population,
    given the critical values of z specified, given that this appears to be a two-tailed test.

    What is the conclusion we must make from this ?

    (5) If the critical values of z were not as extreme, we could conclude that there is a significant bias.
    If we performed a one-tailed test, we could conclude that a bias exists to a certain level of confidence.
    For the critical values of z given, are the results significant ?

    Clearly, there is a chance of making misjudgements by interpreting data in these statistical ways.
    Type one and Type two errors explore this.
    First, be sure about parts (1) to (5).

  3. #3
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    Nov 2009

    Re: Statistics help

    The "null hypothesis" is "statistics jargon" for "nothing out of the ordinary happened".
    In this case it means that "there is no significant difference between the results students get at the schools".

    The "alternative hypothesis" is..... the results differ significantly.

    What do we mean by "significantly"??

    We check where on the bell-shaped standard normal curve, or maybe on the cumulative one,
    that the calculated z-value lies on the horizontal z-axis.


    From the graph, 68.26% of the graph lies within 1 standard deviation away from the mean.
    This corresponds to z=1 or -1, since z=0 corresponds to the mean.
    (z is basically the number of standard deviations above or below the mean that the variable is.
    x-mean calculates the variables distance from the mean, then dividing by the standard deviation
    finds out how many standard deviations the variable is from the mean.
    Once normalised, the variable's position can be evaluated).

    From the graph, 2(47.72)% of the graph lies within 2 standard deviations of the mean.
    95% of the graph lies within 1.96 standard deviations of the mean.
    Therefore, if we are performing a two-tailed test as in this case, we are checking whether or not
    the variable lies in the centre 95% of the graph.
    There is a 5% chance of the variable being in either of the two extreme tails.
    As this is pretty unlikely, the alternative hypothesis will be accepted only if our z reading is out there.
    The reading is saying the following.....

    If the variable value is reflective of the population (no difference between the schools performance levels),
    then there is only a 5% chance of z being >1.96 or <-1.96.
    Hence, if z is between those limits, the null hypothesis will be accepted at the 5% or 0.05 level of significance.

    As z[sub:168b6e61]obt[/sub:168b6e61] is between these limits, it's accepted as being a population reading
    and so it is accepted that the schools do not differ for the measured parameter.

    Type 1 and Type 2 errors to follow.
    Attached Images Attached Images

  4. #4
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    Nov 2009

    Re: Statistics help

    Type 1 error

    Imagine that the mean and standard deviation of the results from the private school
    are the same as the mean and standard deviation of the results from the public schools.
    There would be no difference between the distributions.

    However, when we convert the sample mean to Z[sub:m1cxoxsk]obt[/sub:m1cxoxsk] using

    [tex]\cfrac{sample\ mean-population\ mean}{\cfrac{28}{\sqrt{100}}}[/tex]

    this sample mean can vary quite a bit.
    The weakest students could all have been in the sample of 100 students,
    or the strongest students could have been in the 100,
    or various combinations of students between those extremes.

    Certain samples of students could have produced Z[sub:m1cxoxsk]obt[/sub:m1cxoxsk] above 1.96 or below -1.96,
    even though the distributions could well be identical.
    This ought to happen under "normal" circumstances 5% of the time.

    Therefore, there is a 5% chance or 0.05 probability of rejecting a hypothesis that is in fact true,
    by examining a sample in this way at the 5% significance level.

    This is the probability of making a Type 1 error.

    Type 2 error

    In this case, Z[sub:m1cxoxsk]obt[/sub:m1cxoxsk] is between -1.96 and 1.96, hence there is not enough evidence
    to suggest that the private school results differ significantly from the public schools' results,
    if indeed they do differ. Even if it was beyond the critical values, there is still a 5% chance we
    would be wrongly assuming they are different when they could be the same (Type 1 error).

    The private school's performance could be significantly different.
    It may well be that the sample chosen was not truly reflective of the private school's
    overall performance.

    In order to calculate the Type 2 error, we would need to know the mean and standard deviation
    for the public school. We only have the data from a sample of students from the school unfortunately.

    If we knew both distributions, we could examine the overlap between their respective graphs
    on the same z-axis. This involves locating the critical values of z on the public schools' curve
    and discovering the amount of the curve from the private school that lies within the 2.5% to 97.5% region
    of the public school's curve.
    This overlap gives the probability of assuming the distributions do not differ,
    when in actual fact they do.

    If the population standard deviation for both is 28, then we could calculate the Type 2 error,
    if the private school's population mean was 71.3.
    As it cannot be taken that 71.3 is the private school's mean,
    there is no need to calculate the Type 2 error.


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