A product with an annual demand of 1000 units has Co = $25.50 and Ch = $8. The demand exhibits some variability such that the le
ad-time demand follows a normal probability distribution with μ=25 and σ=5. a.What is the recommended order quantity? b. What are the recorder point and safety stock if the firm desires at most a 2% probability of stockout on any given order cycle? c. If a manager sets the reorder point at 30, what is the probability of a stockout on any given order cycle? How many times would you expect a stockout during the year if the reorder point were used?
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Answer:
A. Null and alternative hypothesis:
B. Yes. At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.
P-value = 0.00002
C. As the difference is calculated as (population 1 − population 2), being population 1: groceries and population 2: dinning out, and knowing there is evidence that the true mean difference is positive, we can say that the groceries annual credit card charge is higher than dinning out annual credit card charge.
The point estimate is the sample mean difference d=$840.
The 95% confidence interval for the mean difference between the population means is (490, 1190).
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample has a size n=42.
The sample mean is M=840.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1123.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size are:
This test is a two-tailed test, with 41 degrees of freedom and t=4.848, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.00002) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.
We have to calculate a 95% confidence interval for the mean difference.
The t-value for a 95% confidence interval and 41 degrees of freedom is t=2.02.
The margin of error (MOE) can be calculated as:
Then, the lower and upper bounds of the confidence interval are:
The 95% confidence interval for the mean difference is (490, 1190).