Suppose you were performing a test to determine the average number of hours kids (12 - 14)spend on their phones. Your hypothesis is that they spend more time that they used to in 2010 when the average number of hours was 4 hours per day. Your hypotheses statement would be as follows:

H0: μ=4

H1: μ>4 (Claim)

After collecting your data you have concluded that you will reject the claim.

State the final conclusion for this test addressing the claim.
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1 Answer

The null hypothesis is that μ=4. The alternative hypothesis (the claim) is that μ>4. The test is to see if the statistical evidence is sufficient to accept or reject the null hypothesis. If the test fails to reject the null hypothesis, the alternative hypothesis cannot be accepted within the degree of certainty imposed on the test. The degree of certainty, the confidence level, is often set in advance to 95% (fairly certain), which corresponds to a significance level of 5%. The test statistic in this case is the mean, and we use the difference between the sample mean and the currently accepted mean μ. In this case, we use the 1-tailed t-test (tables or calculator), because we are only considering probability higher than μ, and degrees of freedom=n-1 (where n is the sample size). If the sample size is large the normal distribution and the Z-test applies, and the critical value from the t-test table/calculator or from the normal distribution table/calculator is found by plugging in the deviation from the accepted mean as a multiple of the standard deviation. If the sample mean is not more extreme than this critical value (that is, isn't bigger than the mean by more than the critical value) the null hypothesis cannot be rejected; but if it is more extreme, the null hypothesis is rejected, and the conclusion is that the claim is probably true, within the set confidence level.

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