Q1. The number of points TEDU basketball team scores against Stanford basketball team is normally distributed. Suppose that over the last 10 games between the two teams TEDU scored the points given below: 74, 59, 62, 61, 70, 62, 66, 62, 75, 59 Solve part (d), (e) and (f) given the information on variance in part (b). Note: Information of part b is in the images d) How many previous game points do you have to sample to ensure that the error of the estimate for the mean score would not exceed 30? e) The coach of the TEDU team claims that on the average the point scored against Stanford team is 70. Construct a hypothesis test that can be used to help the coach test his claim. That is, state your null hypothesis and alternative hypothesis in terms of the relevant population parameter. f) Test the hypotheses constructed in part (e) using a significance level of 0.01. State your conclusion also in words

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Q1. The number of points TEDU basketball team scores against Stanford basketball team is
normally distributed. Suppose that over the last 10 games between the two teams TEDU scored the points given below:
74, 59, 62, 61, 70, 62, 66, 62, 75, 59

Solve part (d), (e) and (f) given the information on variance in part (b).

Note: Information of part b is in the images


d) How many previous game points do you have to sample to ensure that the
error of the estimate for the mean score would not exceed 30?
e) The coach of the TEDU team claims that on the average the point scored
against Stanford team is 70. Construct a hypothesis test that can be used to help the
coach test his claim. That is, state your null hypothesis and alternative hypothesis in
terms of the relevant population parameter.
f) Test the hypotheses constructed in part (e) using a significance level of
0.01. State your conclusion also in words

 

Part (b).
Given that, the variance of the distribution is 25.
V(X)=25
S. D=V/25
=5
Use the equation C.I = X ±tº to construct the interval.
C.I = 65 ± 2. 26 ×
C.I = 65 ± 5. 65
Therefore3, the upper and lower limits of the confidence interval is 65+5.65 and 65-5.65.
Transcribed Image Text:Part (b). Given that, the variance of the distribution is 25. V(X)=25 S. D=V/25 =5 Use the equation C.I = X ±tº to construct the interval. C.I = 65 ± 2. 26 × C.I = 65 ± 5. 65 Therefore3, the upper and lower limits of the confidence interval is 65+5.65 and 65-5.65.
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