statistics

A student claims the average amount of time spent studying for an exam is less than or equal to 120 minutes. A random sample of 101 students is drawn from a normal distribution. The sample mean is 140 with a sample standard deviation of 20. Test the claim with a .01 level of significance and write your conclusion

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asked by Matt
  1. Z = (mean1 - mean2)/standard error (SE) of difference between means

    SEdiff = √(SEmean1^2 + SEmean2^2)

    SEm = SD/√n

    If only one SD is provided, you can use just that to determine SEdiff.

    Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.

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    posted by PsyDAG

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