Last year Marion Dairies decided to enter the yogurt market, and it began cautiously by producing, distributing, and marketing a single flavor—a blueberry-flavored yogurt that it calls Blugurt. The company’s initial venture into the yogurt market has been very successful; sales of Blugurt are higher than expected, and consumers’ ratings of the product have a mean of 80 and a standard deviation of 25 on a 100-point scale for which 100 is the most favorable score and zero is the least favorable score. Past experience has also shown Marion Dairies that a consumer who rates one of its products with a score greater than 75 on this scale will consider purchasing the product, and a score of 75 or less indicates the consumer will not consider purchasing the product.
Emboldened by the success and popularity of its blueberry-flavored yogurt, Marion Dairies management is now considering the introduction of a second flavor. Marion’s marketing department is pressing to extend the product line through the introduction of a strawberry-flavored yogurt that would be called Strawgurt, but senior managers are concerned about whether or not Strawgurt will increase Marion’s market share by appealing to potential customers who do not like Blugurt. That is, the goal in offering the new product is to increase Marion’s market share rather than cannibalize existing sales of Blugurt. The marketing department has proposed giving tastes of both Blugurt and Strawgurt to a simple random sample of 50 customers and asking each of them to rate the two flavors of yogurt on the 100-point scale. If the mean score given to Blugurt by this sample of consumers is 75 or less, Marion’s senior management believes the sample can be used to assess whether Strawgurt will appeal to potential customers who do not like Blugurt.
Prepare a managerial report that addresses the following issues.
- Calculate the probability the mean score of Blugurt given by the simple random sample of Marion Dairies customers will be 75 or less.
- If the Marketing Department increases the sample size to 150, what is the probability the mean score of Blugurt given by the simple random sample of Marion Dairies customers will be 75 or less?
- Explain to Marion Dairies senior management why the probability that the mean score of Blugurt for a random sample of Marion Dairies customers will be 75 or less is different for samples of 50 and 150 Marion Dairies customers.
Initial post prompt: Your Managerial Report serves as your initial post to the discussion forum. After responding to the requirements posed by the Managerial Report, also provide an example in your career in which you believe one of the lessons learned from the Case has been/could be applicable. Alternatively, if you don’t have/foresee direct experience relevant to your current position, what type of scenario can you anticipate occurring where you can utilize one of the lessons learned from examining this case?
Response post prompt: Consider Managerial Reports posted by two of your peers. One or both of your responses may be to Managerial Reports for a case problem different from your own. Think critically and ask open-ended questions. If you agree, consider their position and expand upon their ideas. Provide an additional perspective. If you disagree, provide your reasoning. Always be professional and courteous in your responses.
Post by classmate 1
I find it interesting that the senior management at Marion Dairies has chosen to have their sample be defined by someone who doesn’t like their blueberry yogurt. I understand that they want to expand their market; however, the logic they are choosing seems flawed and likely to cause nonsampling error because negative opinion of one flavor or one product may bias the result of the strawberry sampling.
Another potential for nonsampling error that I noticed was surrounding the names of the yogurt flavors. In my opinion the name Blugurt was cute for about two seconds, and then it seems to be something I do not want associated with my food which has the potential to become a marketing nightmare. The name Strawgurt is even worse. I would strongly suggest the sampling to be a blind study where the subjects of the sample are not aware of the names of the yogurt because that could possibly cause unintended nonsampling error if the subject has a particular opinion regarding the name of the yogurt. According to our text, nonsampling error can introduce bias into the sample results which can ultimately mislead the decision makers (Anderson et al, 2020).
Alas, that is not what we are here to talk about this evening! First we are asked to calculate the probability that the mean score of Blugurt given by a simple random sample will be 75 or less on a scale of 0 to 100. In order to calculate this probability, I calculated the z value as -1.4142 and used the standard normal distribution table to find the probability of 0.07865. I did the same with an increased sample size of 150 and found the probability to be 0.007153.
While the mean remains the same, the increase in sample size causes the standard error to decrease thus impacting the probabilities. Our text also mentioned that the decrease in standard error suggests that the data from a larger sample size becomes more reliable.
Anderson, D., Williams, T., Cochran, J., Sweeney, D., Camm, J., Fry, M., & OhlmannB, J. (2020). Modern business statistics with Microsoft Excel (7th ed.). Boston, MA: Cengage.
Post by classmate 2
For this week’s discussion, I chose case problem 1 from chapter 7: Marion Dairies. In this case study, Marion Dairies’ management and marketing team want to analyze a sample of 50 customers to gain insight on the success of their new yogurt flavor, Blugurt. Customers were asked to rank the flavor on a 100-point scale (0 being least favorable and 100 being the most favorable). A score of 75 or less is believed to indicate that a consumer will not consider purchasing the product. For this case, we are provided with a mean score of 80, a standard deviation of 25, and two varying sample sizes to compare (50 and 150). For this simple random sample, we look to analyze the distribution of the sample mean, specifically the probability of the mean occurring at or below a customer rank of 75. Because the population of customers purchasing yogurt is infinite, we can dismiss the application of the population correction factor and simplify the calculation of the standard error to the standard deviation (25) divided by the square root of the sample size (either 50 or 150) (Anderson 2021). Furthermore, given that the sample sizes both exceed 30, general statistical practice ensures that we are observing a normal or near normal distribution, as implied by the rules of the Central Limit Theorem (Anderson 2021). Below is a summary of findings outlined in the managerial report synopsis. Attached is an Excel file to illustrate the findings.
- Calculate the probability the mean score of Blugurt given by the simple random sample of Marion Dairies customers will be 75 or less:
The probability of the sample mean being 75 or less for Blugurt with a sample size of 50 is 0.07865 or approximately 7.9%. This is demonstrated in the first tab of the attached Excel file and is represented by finding the z value from the population sample of 50. This is calculated by first finding the standard error, represented by the standard deviation divided by the square root of the sample population (3.536) (Anderson 2021). Then the z value is found by subtracting the mean from the target value of 75 and dividing that sum by the standard error (-1.414) (Anderson 2021). By obtaining the z value, excel function (NORM.DIST) can be used to calculate the cumulative probability (area to the left of z = -1.414) which provides the probability that the sample mean will be less than 75 (0.07865 or approx. 7.9%) (Anderson 2021).
- If the Marketing Department increases the sample size to 150, what is the probability the mean score of Blugurt given by the simple random sample of Marion Dairies customers will be 75 or less?:
The probability of the sample mean being 75 or less for Blugurt with a sample size of 150 is 0.00715 or approximately 0.7%. This is demonstrated in the second tab of the attached Excel file and is calculated in the same way as before. The standard error (2.041) from the standard deviation divided by the square root of the sample population, then the z value by subtracting the mean from the target value of 75 and dividing that sum by the standard error (-2.449) (Anderson 2021). Similarly, by obtaining the z value, excel function (NORM.DIST) is used to calculate the cumulative probability (area to the left of z = -2.449) which provides the probability that the sample mean will be less than 75 for a sample of 150 (0.00715 or appox. 0.7%) (Anderson 2021).
- Explain to Marion Dairies senior management why the probability that the mean score of Blugurt for a random sample of Marion Dairies customers will be 75 or less is different for samples of 50 and 150 Marion Dairies customers:
The reason that the probability varies with the sample size is because the standard error of the mean is affected by changes to the sample size (Anderson 2021). Whenever the sample size is increased, we are accounting for more data and therefore receiving values with less variation and that are more closely representing the population mean (Anderson 2021). As demonstrated above, the sample size increase reduces the standard error and therefore reduces variation in the values collected. Given that the probability and variability decreased with a larger sample, it is safe to assume that there is evidence that the population favors Blugurt and is likely to purchase this product. A similar simple random sample would also be beneficial for gathering point estimations and statistical inference on customers’ views on Strawgurt. By comparing the data collected on both flavors, management can evaluate preference for one or the other and potentially identify if Strawgurt offers another favorable option for customers and hence strengthens their market share, or if it would indeed cannibalize the existing favor for Blugurt and have offsetting outcomes.
I personally have not used sample probabilities to make statistical inference on larger populations however, I think that this is a useful tool and concept to understand so that it can be applied to a situation where information is needed for a population that is too large to fully observe. Patient satisfaction is an important metric in evaluating our clinic’s likelihood to recommend and this would be a useful tool to experiment with when we offer a new service such as telemedicine or reservation options within our service line. Since the entire population cannot be surveyed, data gathered on a random sample could provide meaningful insight into the success of changes to our business model.
Anderson, D. R., Sweeney, D. J., Williams, T. A., Camm, J. D., Cochran, J. J., (2021). Modern Business Statistics with Microsoft Office Excel (7th ed.)