Week 6 Statistics Qnt 561
H0: game length is >= 3.5 hours
Ha: game length is < 3.5 hours
mean = 2.9553 stdev = 0.5596
Get the t test statistic: t = (xmu)/(stdev/sqrt(N)) t = (2.95533.5)/(0.5596/sqrt(17)) t = 4.0133
Get the critical value for df = N1 = 16, one tail, alpha is 0.05:
1.7459
Since our test statistic is much lower than the critical value, we reject the null hypothesis. There is enough evidence to conclude that games are shorter than 3.50 hours.
Chapt 11 # 58
The amount of income spent on housing is an important component of the cost of living. The total costs of housing for homeowners might include mortgage payments, property taxes, and utility costs (water, heat, electricity). An economist selected a sample of 20 homeowners in …show more content…
b. Is there a difference in the cars? No.
Chapt 13 # 37
H0: correlation 0
Critical value, from a t table, one tail, with df = n2 = 23:
1.7139
Test statistic:
r*sqrt((n2)/(1r^2))
0.94*sqrt(23/(10.94^2))
= 13.213
Chapt 13 # 40
Chapt 14 # 17
a. What are the estimated sales for the Bryne store, which has four competitors, a regional population of 0.4 (400,000), and advertising expense of 30 ($30,000)?
Use the regression:
14  1*4 + 30*0.4 + 0.2*30
= 28
So the sales would be $28000, since the regression gives the value in thousands
b. Compute the value of r^2. ssr/sst = 3050/5250
= 0.58095
c. Compute the multiple standard error of estimate. sqrt(mse) = sqrt(84.62)
= 9.1989
d. Conduct a global test of hypothesis to determine whether any of the regression coefficients are not equal to zero. Use the .05 level of significance.
H0: coeffs are zero
Ha: coeffs are not zero
F = 1016.67/84.62
= 12.01454
The critical value, from an F table, with df = 3, 26 is: 2.975
The test statistic is much higher, so we reject the null.
At least one coefficient is non zero.
e. Conduct tests of hypotheses to determine which of the independent variables have significant regression coefficien
Each hypothesis looks like this:
H0: coefficient for Xn is 0
Ha: coefficient for Xn is non zero
The critical T value, from a table, is 2.056
The t value for X1 is not outside that range, so we don't reject.
The t value for X2 and X3 are outside, so

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