The current demand forecasting method is based on qualitative techniques more than quantitative ones. If the forecast is not accurate, the company would carry both inventory and stock out costs. It might lose customers due to shortage of supply or carry additional holding costs due to excess production. If the actual demand doesn’t match the forecast ones, and the forecast was too high, this will result in high inventories, obsolescence, asset disposals, and increased carrying costs. When a forecast is too low, the customer resorts to a competitive product or retailer. A supplier could lose both sales and shelf space at that retail location forever if their predictions continue to be inaccurate. The tolerance level of the average consumer …show more content…
Thus we reject Ho and t is significant.
For d1, t-statistic=- 2.5334, t-statistic > t-critical. Thus we reject Ho and d1 is significant.
For d2, t-statistic= 1.8774, t-statistic < t-critical. Thus we do not reject Ho and d2 is not significant.
For d3, t-statistic= 8.8773, t-statistic > t-critical. Thus we reject Ho and d3 is significant.
This shows that there is a time trend and seasonality in the quantity demanded of PVB.
Therefore, the equation for PVB forecast is the following:
Q= 32561.2+ 1977.6t - 13193.1D1 +9631.5D2 + 45122.1D3
Forecasting for Fire Valves:
The past demand pattern of Fire Valves, different than PVB’s, reveals random fluctuations and does not seem to be affected by seasonality factors. This is why we cannot rely on past data to forecast future demand. Given that Barge and Connors released the new fixed-pressure valves, forecasting demand for fire valves is not of great value anymore. This new product is expected to have a dramatic growth. Therefore, analyzing the sales of fire valves is irrelevant. Instead, in order for Barge to forecast the demand for the new product, he can make a study using the unemployment rate, bank prime loan rates, and number of new housing starts. This study cannot give
The critical value for for this two-tailed test is. The rejection region is given by
From the above output, we can see that the p-value is 0.000186, which is smaller than 0.05 (if we select a 0.05 significance level).
15 In testing the hypotheses: H0 β1 ’ 0: vs. H1: β 1 ≠ 0 , the following statistics are available: n = 10, b0 = 1.8, b1 = 2.45, and Sb1= 1.20. The value of the test statistic is:
We reject Ho if χ2 > χα2. At α=0.05, with 4 degrees of freedom, the critical value becomes χα2=9.488 (table E.4)
The null hypothesis is rejected since the p-value is below the significance level of 0.05.
In this study, t= -3.15 describes the mental health variable. It is significant because they are the variables being tested since the p value is 0.002 and the alpha is 0.05, the difference can cause the null hypothesis to be rejected.
Consider the following scenario in answering questions 5 through 7. In an article appearing in Today’s Health a writer states that the average number of calories in a serving of popcorn is 75. To determine if the average number of calories in a serving of popcorn is different from 75, a nutritionist selected a random sample of 20 servings of popcorn and computed the sample mean number of calories per serving to be 78 with a sample standard deviation of 7.
One-sample t-test are used in the parametric test which analyzes the means of populations. The t-test for independent groups are statistics that relates difference between treatment means to the amount of variability expected between any two samples of data within the same population (Hansen & Myers, 2012). Critical values are used in significant testing provide a range of t distribution that is used in whether a null hypothesis is rejected. Based on the data below as the level of significance is at .05, thus the critical values would fall under ±1.860 and the t value for this is 1.871 would suggest for the null to be rejected as it is greater than the critical value (Privitera, 2015, p. 267). Based on the population mean of 70 there was a mean difference of
The next table shows the results of this independent t-test. At the .05 significance level, can we conclude
H0: μ ≥ 3200 against H1: μ < -1.7823 when I rejected the null hypothesis.
located down the row of the t-table. The critical value result is the point where the a-level and degrees of
Since 3.27 the t statistic is in the rejection area to the right of =1.701, the level of
5) From calculations, computed z value is more than -1.65 and falls within Ho not rejected region. Ho is not rejected at α = 0.05 & α = 0.01 significance levels.
There are two hypotheses that were tested in relation to the overall fit of the model:
The result of heteroskedasticity test done for Model II is also shown in Table 5.13 below. The null hypothesis of no heteroskedasticity cannot be rejected this time too. That means the standard errors, T-statistics and F-statistics done for the model are valid.