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Math Help - Statistics - Hypothesis testing

  1. #1
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    Statistics - Hypothesis testing

    hi, i am having problems currently with these two problems, attached are pictures of the two problems i am having problems with, i dont know what kind of testing i am supposed to use. Also how do i find the p-values? My professor wants me to do these problems in a specific format attached is a rubric on how we should show our work.



    here is the first problem



    second



    thanks in advance for the help
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by kaistylez View Post

    here is the first problem


    What is SRS, presumably some sort of random sample, but what sort?

    Simple Random Sample, Stratified Random Sample, ...?

    RonL
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  3. #3
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    srs is a simple random sample, tahnks for the reply
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  4. #4
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    Problem 4, a
    This problem is a hypothesis test for the mean of a large sample.

    First we identify the null and alternative hypotheses and label the given information.

    H_0: \mu _{increase} \le 0 Null Hypothesis
    H_1: \mu _{increase} > 0 Alternative Hypothesis (Claim)

    n=200
    \bar{x}_{increase}=332
    s=108
    \alpha=1\%=0.01

    We assume that the null hypothesis is true and choose to reject it if P\le\alpha and fail to reject it otherwise.

    To find the P-value, we first find the z-score of the test statistic and then get the area bounded to its right.

    z_{\bar{x}}=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}} = \frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}} by the Central Limit Theorem
    =\frac{332-0}{\frac{108}{\sqrt{200}}} \approx \frac{332}{7.34} \approx 43

    Because the alternative hypothesis contains the greater than inequality symbol, the hypothesis test is a right-tailed test and P\approx 0.

    P<\alpha so we reject the null hypothesis. Therefore, the alternative hypothesis H_1 is true and ommitting the annual credit card fee for customers who charge at least $2,400 in a year WILL increase the amount charged on the bank's credit cards.

    Problem 4, b
    We know that the c-confidence interval for the population mean of increase is:
    \bar{x} - E < \mu < \bar{x} + E
    \bar{x} - z_c\frac{s}{\sqrt{n}} < \mu < \bar{x} + z_c\frac{s}{\sqrt{n}}

    So substituting the given values gives us:
    332 - \frac{2.575\cdot108}{\sqrt{200}} < \mu < 332 + \frac{2.575\cdot108}{\sqrt{200}}
    332 - \frac{278.1}{\sqrt{200}} < \mu < 332 + \frac{278.1}{\sqrt{200}}
    332 - 19.66 < \mu < 332 + 19.66
    312 < \mu < 352

    ----------

    Although it all seems to make sense, the z-score of 43 is ridiculous so I am not completely sure about the correct answer. This is the best I could do... hope it helps.
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