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Given a sample size of 38, with sample mean 660.3 and sample standard deviation 95.9 we are to perform the following hypothesis test.
Null Hypothesis H0: ƒÊ = 700
Alternative Hypothesis H0: ƒÊ ‚ 700
At significance level 0.05

3. Calculate the test statistics
(Tip: this is the case when we test the claim about population mean with population standard deviation not known; 95.9 is a sample standard deviation not a population standard deviation).
4. Use Table A-3 to find critical value for this test and make decision:
reject or do not reject the Null Hypothesis?

  • statistics - ,

    Usually you use z-tests when sample sizes are large (n is greater than
    or equal to 30) whether or not you know the population standard deviation.

    If you do not know the population standard deviation and have a small
    sample (n < 30), then you can use t-tests.

    Therefore, try a z-test for your data.
    Using the z-test formula to find the test statistic:
    z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)

    With your data:
    z = (660.3 - 700)/(95.9/√38)

    I'll let you finish the calculation.

    Determine the critical value using your table. Determine if this is a one-tailed or two-tailed test. (Hint: The test is one-tailed if the alternative hypothesis shows a specific direction; the test is two-tailed if the alternative hypothesis does not show a specific direction.) Once you have the critical value, compare it to the test statistic calculated above. If the test statistic exceeds the critical value, reject the null. If the test statistic does not exceed the critical value, then do not reject the null.

    I hope this will help get you started.

  • statistics - ,

    I do very much appreciate your help. This guides me to where I need to go.

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