Economics Math Problem Week 4 Practice Worksheet (Math Problem Sample)

Week 4 Practice Worksheet
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Week 4 Practice Worksheet
Question 1: Can one conclude that there is a difference in the mean times of the Prada and the Oracle at the 0.5 significance point? Elucidate the results to an individual familiar with a single sample test but unfamiliar with the independent mean t- value.
The problem has two values, and they are assigned as follows;
X – The time for Prada in minutes
Y – The time for Oracle in minutes
However, we have to assume that the variables are independent and with uneven variances. Thus, the interest at this juncture is to test the difference in the averages of the two and thus;
X~N (u1, sigma 12) whereas Y ~N (u2, Sigma 22)
(H0) First/ null hypothesis u1 – u2 – 0
(H1) second/ alternative hypothesis: u1 – u2 ≠ 0
For a single sample test, there is only one variable that can be assigned the value (S) and an (n) sample size whose variance is unknown. The unknown mean can be presumed to be u0 for H0
The sample mean = S-bar
Sample standard deviation = s
Therefore;
T= (S-bar-u0) sqrt (n)/s that follows a T- distribution under h0 and the df n-1
The above is a single sample test
However, there are two independent samples; X~N (u1, Sigma 12) and Y ~N (u2, Sigma 22). Besides, the sample size is n1 from X and n2 from the value Y. As such, the X-Y ought to follow a standard distribution with the mean u1 – u2 and the variance of Sigma 22 = Sigma 12
Assume that the Y-bar and X-bar are the sample means whereas S2 and S1 are their respective sample standard deviations Following the fact that Sigma 1 and Sigma 2 are unknown they are consequently estimated by S1and S2 respectively. Thus T= {(X-bar- Y-bar)-(u1 – u2)} sqrt(S12/n1 + S22/n2) follow a distribution under H0 u1-u2=0
Therefore, T= {(X-bar- Y-bar) 1/sqrt (S12/n1 + S22/n2) and follows a T-distribution on H0
As such, they are the two independent sample t evaluations from the sample provided that X-bar=12.170, Y-bar=14.875, S1 = 1.056, S2= 2.208, N1 =10 and N2=12. Thus the value of T is 3.76. The alternative hypothesis has two sides and thus rendering the test to be a two- tailed hypothesis
P value = p=2*min {P (T>-3.76)P(T<-3.76)} in the event where T follows a T-distribution with a
Df from p=0.0017
The significance level is alpha =0.05, and therefore p is less than alpha
Therefore, the H0 is rejected purporting that there exists a difference in their means.
Question 2: Is there a variation in the mean quantity procured on impulse by the mentioned stores? Explain these outcomes to an individual familiar with the single t test sample for a single sample but unfamiliar with the independent means for t test
PlumPeachAverage 18.292115.8680Standard Deviation2.55272.3306n1410df20Difference -2.41Difference standard error1.00431Hypothesized difference0t-2.41Two tailed p value0.0255Lower 99% confidence interval-5.28173Upper 99% confidence Interval0.43345Error margin2.85759
The average value for the plum street is 18.22921 whereas that of Peach Street can be said to be 15.8680. On the other hand, their standard deviation is 2.5527 and 2.3306 for the plum and peach streets respectively. The n-value is 14 and 10.
H0: µ plum = µ peach
H1: µ plum в‰  µ peach
The hypothesis is to be rejected if t > 2.819
Therefore, t= 2.3530
The hypothesis is not to be rejected since there is no difference in their mean values
Question 3: Is there a difference in the average daily calls for the two workers? Calculate the P- value.
Hypothesis test for the two groups (pooled variance and T-test)
4.77 – Larry and 5.02 for George (Average)
1.05 – Larry and 1.23 for George (Standard Deviation)
40 – Larry and 50 for George (n)
The difference between the two is -0.25 (Larry’s valu...