1. A company claims that the mean weight per apple they ship is 120 grams with a standard deviation of 12 grams. Data generated from a sample of 49 apples randomly selected from a shipment indicated a mean weight of 122.5 grams per apple. Is there sufficient evidence to reject the company’s claim? (Alpha = .05)
What is given?
(lower case sigma) = population standard deviation = 12 grams
SRS (simple random sample) mean of 122.5 g, n= 49
TEST OF SIGNIFICANCE TEMPLATE
1. PARAMETER OF INTEREST
Is there sufficient evidence to reject the company’s claim that the mean weight of apples is in fact equivalent to mu = 120 grams.
2. CHOICE OF TEST
Only have one sample and one population and therefore you are also given the st dev so use the one-sample z-test for a mean. Because n > or equal to 30 at 49 it is considered a large sample.
3. CHECK OF ASSUMPTIONS
n>49 do you know the pop dev (lower case sigma), testing to prove the mean. One sample. good
4. NULL HYPOTHESIS
The values are equal or there is no difference or change in the mean. There is no difference between the weight of the sample and the population. Mu = 120
5. ALTERNATIVE HYPOTHESIS
The mean weight is NOT the same between the sample and the population. Mu is not equal to the sample mean.
6. PROBABILITY STATEMENT
Alpha is .05 and if P is less than .05 we Reject the Null
7. TEST STATISTIC: Use the formula one sample z-test
Look at the chart to find the p –region which w numerically found below and determine of the alpha falls within it or outside.
9. P VALUE
P=.0723 times 2 areas which means because it is two-tailed overall p – value is
.1446 > .05 we fail to reject the null
10. RECOMMENDED DECISIONS
In conclusion, the alpha value if less than the p value based on the z-score provided and therefore we fail to reject the null.
11. INTERPRETATION
There isn’t sufficient evidence at the .05 level of significance to reject the company claim that the mean weight per apple in the shipment is 120 grams.
