Posted by tanya on .
An article claimed that the typical supermarket trip takes a mean of 22 minutes. suppose that in an effort to test this claim, you select a sample of 50 shoppers at a local supermarket. the mean shopping time for the sample of 50 shoppers is 25.36 min. with a standard deviation of 7.24 minutes. using a 0.10 level of significance is there evidence that the mean shopping time at the local supermarket is different from the claimed value of 22 minutes?
***if i plug everything into Z= mean22minutes/ standard deviation over sqrt of n is that the answer? i am soo confuseddd

Statistics...help 
MathGuru,
Let me give you a little background and maybe this will help you understand a few of the concepts of hypothesis testing.
We use samples to support some hypothesis about a population. We make inferences about the population from sample data. When setting up a statistical test, we form two hypotheses: a null hypothesis and an alternative hypothesis. The null hypothesis tests the probability that the sample mean is very likely drawn from the population mean. If there is a low probability that the sample mean has been drawn from the population mean, we will reject the null hypothesis and accept the alternative hypothesis. The alpha level (also called the significance level) is the probability at which we reject the null and accept the alternative hypothesis. For example, if the alpha level is set at 0.05, there is a 5% probability that we will reject the null hypothesis. This is a very common alpha level used in statistics. Often we use tables to look for values that correspond to the alpha level stated. These values are known as critical or cutoff values. They cut off the distribution at that point. If we have an observed value (calculated from a formula) that exceeds our critical value or cutoff value from a table, then we have to reject the null hypothesis and accept the alternative hypothesis. If the observed value does not exceed the critical or cutoff value from a table, then we fail to reject the null hypothesis. How do we translate 0.05 into a critical value? It depends on the type of test we are doing. For a onetailed test, we don't split the value. For a twotailed test, we split the .05 into .025 and .025 for both tails of the distribution curve (a twotailed test is like a confidence interval in that respect). If we use a ztable for a onetailed test at .05 level of significance (meaning the alternative hypothesis is showing a specific direction like "less than" or "greater than" in its statement), then we will have a critical or cutoff value of z = 1.645 or it could be z = 1.645, depending on the direction. This will determine where we "draw the line" to reject the null hypothesis. If we use a ztable for a twotailed test at .05 level of significance (meaning the alternative hypothesis is showing no specific direction and uses "does not equal" in its statement), then we will have a critical or cutoff value of z = + or  1.96 (meaning either tail of the distribution curve). If we had a test result of z = +2.00 (for example), then we would have exceeded the positive critical value of +1.96 for this particular test result and the null hypothesis would be rejected in favor of the alternative hypothesis.
Now let's look at your problem.
You need to set up hypotheses, calculate the ztest statistic (since this is a ztest), then compare to the critical value from a ztable to determine whether or not to reject the null hypothesis.
Hypotheses:
Ho: µ = 22 >this is the null hypothesis.
Ha: µ does not equal 22 >this is the alternate or alternative hypothesis.
This would be a twotailed or nondirectional test because the alternative hypothesis doesn't specify a specific direction.
The reason we know this is a twotailed test is because the problem asks if there is evidence that the mean time is different, which means the results could be in either tail of the distribution.
Therefore, using a ztest formula:
z = (sample mean  population mean)/(standard deviation divided by the square root of the sample size)
With your data:
z = (25.36  22)/(7.24/√50)
Finish the calculation.
Now you will need to find the critical value at 0.10 level of significance using a ztable. Since this is a twotailed test, we split the 0.10 into 0.05 and 0.05 for both tails of the distribution curve. Find the critical value, then compare to your test value. Determine whether or not to reject the null.
I hope this explanation will help.