It is an exam. I will send your all of my lecture notes.The mid-term exam has 4 questions all problem solving, any calculations done in Excel is acceptable just make sure you attach your excel file...

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It is an exam. I will send your all of my lecture notes.The mid-term exam has 4 questions all problem solving, any calculations done in Excel is acceptable just make sure you attach your excel file along with your exam.


The following table shows the | 1 prices and quantities | consumed by a family of four | for selected food commodities for 1990 and 2006. Use 990 as the base period. a. Determine the simple index for steak for 2006. 1990 2006 | Price | Quantity | Price | Quantity 15139 30 $2.09 110 11.69] 100 2.75 120 | 1.29 95 2.59 100 1 a99 50 9.49 40 b. Determine the simple aggregate price index for 2006. ¢. Determine the Laspeyres price index for 2006. a. d. Determine the Paasche price index for 2006. 9. 10. A firm sold $50.000 of a particular product in 1995 and $120,000 in 2003. Using 1995 as the base, what is the index for 2001? a. $70,000 b. 240 percent c. 140 percent. d. 41.7 percent An index has 1980 as its base. The index reported in 1993 was 127.2 and in 2003 it was 186.7. The percent increase from 1993 to 2005 is a. 46.8 percent. b. 27.2 percent. c. 59.5 percent. d. none of the above. Which of the following price indexes uses current period quantities in its base? a. a value index b. a simple index c. Laspeyres Price Index d. Paasche Price Index Part II: Record your answer in the space provided. Show essential work. 11. The ecarnings per share for General Year | Earnings per share Index Electric from the 2002 Annual Reportare [190g $0.95 given at the right. Develop an index 1900 700 showing the change in earnings for the - given years. Use 1998 as the base period. | 2000 oo 1.29 2000 | 1.41 | 2002 1.51 [2003 | 1.52 2004 1.62 [2005 | 1.55 12. Professor Jim Martin had an annual income in the base period of $30,000. In YE 12 2005 his annual income was $75,100. During the same period the CPI rose from 100 to 184.6. What was his real income in 2005? | 14. e. Determine a value index for 2006. The table reports the net profit for Heban Tool and Die, Inc. for | Year the years 1996 and 2006. Also reported is the tool and die index [1996 for the same years (1985 = 100). 2006 a. What was the percent increase in the index from 1996 to 2006? b. Convert the index to a 1996 base. What is the new index for 2006? o Net profit | Index $45,380 | 136.3 65,035 | 150.2 a. b. c. Determine the net profit for 2006 in terms of the 1996 base. Comment on the change.
Answered 3 days AfterMay 11, 2023

Answer To: It is an exam. I will send your all of my lecture notes.The mid-term exam has 4 questions all...

Prithwijit answered on May 14 2023
27 Votes
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1.a. Let the GPA scores of the sophomore be denoted by X1 and for the juniors let it be X2.
Let the mean GPA score of sophomores be and for the GPA s
core of the juniors be
So, according to the questions, we have to test whether the mean GPA of sophomore and junior are equal or the mean GPA of sophomore is less than the junior. Mathematically,
H0: = vs H1:<
b. Here the test statistic follows a t-distribution with 28 degrees of freedom. The test is a left-hand test, so the critical point(t28,0.05) for 5% level of significance is defined as
                P(TWhere T is the corresponding test statistic which follows a t-distribution with 28 df.
The above is the formula we used in excel and got the critical point t28,0.05 = -1.70113
c. Since the test is a left-hand test, therefore the H0 is accepted or rejected is solely based on the small cut-off values. For this particular test the decision rule or the critical region is defined as –
            T < t28,0.05, accept the null hypothesis(H0)
            T > t28,0.05, reject the null hypothesis(H0)
d. d. The test statistic is defined as –
        T =
here S = pooled variance of sophomore and junior
n1 = sample size of size for sophomore
n2 = sample size of size for junior
under H0 the T follows t-distribution with (n1+n2 – 2) degrees of freedom
e. So, T =
= -0.86
Since observed T is greater than the critical value, hence the null hypothesis is accepted.
f. The final conclusion is that the there is no significant difference in the mean GPA of...
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