. A company has 48 sales representatives, one for each state in the contiguous U

Question

. A company has 48 sales representatives, one for each state in the contiguous United States. Each state is divided into four regions. Each state sales representative only reports the average sales from the state he or she covers each week. The average reported is calculated from one individual data point per region. Before the New National Sales Manager was hired, the average sales reported per state each week was $95,000 with a standard deviation of $12,000. The National Sales Manager claims that under her leadership, sales have increased despite the weakening economy. A sample was taken when the manager’s annual review was being written and average sales were $105,100. a) Based on an , can the National Sales Manager make her claim? Use and show the 5-Step Process. b) What is the cutoff point of sales that allows or would allow the National Sales Manager to make the claim?

Explanation / Answer

a) Set up your null and alternative hypotheses. The null hypothesis should generally be that there was no change in the variable of interest. The alternative should be the claim you are testing.

Let be the true average weekly sales.

We would expect the distribution to be approximately normal, so we'll use a normal distribution to model the data.

What we're really interested in now is the probability that we could have gotten a sample mean of $105,100 if the true mean is still 95,000. If it turns out we are very unlikely to get a number as high as 105,100 when the true mean is 95,000, then that the true mean is probably greater than 95,000, which is evidence to support our claim that sales have increased under the new manager.

We need to figure out where this 105,100 number falls on our distribution, so we've gotta find its Z score (standard score).

Then we can use the normal cumulative density function (normalcdf) or a Z table to find the probability that Z > 0.8417 on a standard normal distribution.

Normalcdf takes two parameters: the lower bound z score and the upper bound z score, and gives you the probability area between them.

We want to find P(Z > 0.8471), so do:

The 100 is just an upper bound... something really high so that we're effectively saying "give me the probability that Z > 0.8471".

This yeilds a probability, called your p-value. You can interpret this value as "The chance that we could get a sample mean as extreme as $105,100 if the true population mean is $95,000."

Now we have to figure out whether there's evidence to support the claim at the confidence/alpha level we chose (=0.02).

If your p-value is greater than 0.02, you say that you "fail to reject the null hypothesis". That means that the probability that we could have gotten a number like $105,100 due to chance is too high, and that there isn't evidence to support the manager's claim at the 2% confidence level.

If your p-value is less than 0.02, you say that you "reject the null hypothesis in favor of the alternative". Basically, that the chance of getting a sample mean as extreme as $105,000 is low enough that there is evidence to support the claim that the mean is actually greater than 95,000 (at the 2% confidence level).

b) You have to find a Z score that corresponds to the 98th percentile. Use inverse norm of 0.98 ( invnorm(0.98) ).

2.057 = (x - 95000)/12000

x = 119684

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