In an early home game, an NBA team made 66.32 percent of their 95 free throw att
ID: 3233062 • Letter: I
Question
In an early home game, an NBA team made 66.32 percent of their 95 free throw attempts. In one of their last home games, the team had a free throw percentage equal to 73.86 percent out of 88 attempts. Do basketball teams improve their free throw percentage as their season progresses? Test the hypothesis of equal free throw percentages, treating the early season and late season games as random samples. Use a level of significance of .10.
7.
value:
10.00 points
Required information
Assume 1 is the early proportion of free throws and 2 is the later proportion. Choose the correct null and alternative hypotheses.
H0: 1 – 2 0 vs. H1: 1 – 2 > 0
H0: 1 – 2 0 vs. H1: 1 – 2 < 0
H0: 1 – 2 = 0 vs. H1: 1 – 2 0
References
eBook & Resources
Multiple ChoiceLearning Objective: 10-05 Perform a test to compare two proportions using z.
Difficulty: 2-MediumLearning Objective: 10-07 Use Excel to determine p-values for two-sample tests using z or t.
Check my work
8.
value:
10.00 points
Required information
References
eBook & Resources
WorksheetLearning Objective: 10-05 Perform a test to compare two proportions using z.
Difficulty: 2-MediumLearning Objective: 10-07 Use Excel to determine p-values for two-sample tests using z or t.
Check my work
9.
value:
10.00 points
Required information
Calculate the Z test statistic value. (Round your answer to 4 decimal places. Negative value should be indicated by a minus sign.)
References
eBook & Resources
WorksheetLearning Objective: 10-05 Perform a test to compare two proportions using z.
Difficulty: 2-MediumLearning Objective: 10-07 Use Excel to determine p-values for two-sample tests using z or t.
Check my work
10.
value:
10.00 points
Required information
References
eBook & Resources
WorksheetLearning Objective: 10-05 Perform a test to compare two proportions using z.
Difficulty: 2-MediumLearning Objective: 10-07 Use Excel to determine p-values for two-sample tests using z or t.
Check my work
11.
value:
10.00 points
Required information
References
eBook & Resources
WorksheetLearning Objective: 10-05 Perform a test to compare two proportions using z.
Difficulty: 2-MediumLearning Objective: 10-07 Use Excel to determine p-values for two-sample tests using z or t.
Check my work
12.
value:
10.00 points
Required information
References
eBook & Resources
WorksheetLearning Objective: 10-05 Perform a test to compare two proportions using z.
Difficulty: 2-MediumLearning Objective: 10-07 Use Excel to determine p-values for two-sample tests using z or t.
Check my work
13.
value:
10.00 points
Required information
References
eBook & Resources
WorksheetLearning Objective: 10-05 Perform a test to compare two proportions using z.
Difficulty: 2-MediumLearning Objective: 10-07 Use Excel to determine p-values for two-sample tests using z or t.
Check my work
In an early home game, an NBA team made 66.32 percent of their 95 free throw attempts. In one of their last home games, the team had a free throw percentage equal to 73.86 percent out of 88 attempts. Do basketball teams improve their free throw percentage as their season progresses? Test the hypothesis of equal free throw percentages, treating the early season and late season games as random samples. Use a level of significance of .10.
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
1) Null hypothesis: P1 - P2 = 0
Alternative hypothesis: P1 - P2 < 0
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a two-proportion z-test.
Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).
2) zcritical = - 1.28
We will reject the null hypothesis if z value of test is less than - 1.28.
p = (p1 * n1 + p2 * n2) / (n1 + n2)
p = 0.6995
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.06784
3) z = (p1 - p2) / SE
z = - 1.11
Since z value does not lie in the rejection region hence we have to accept the null hypothesis.
where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.
4) Since we have a one-tailed test, the P-value is the probability that the z-score is less than - 1.11. We use the Normal Distribution Calculator to find P(z < - 1.11) = 0.1335
Interpret results. Since the P-value (0.1335) is greater than the significance level (0.10), we cannot reject the null hypothesis.
From the above test we do not have sufficient evidence in the favor of the claim that basketball teams improve their free throw percentage as their season progresses.
They do not improve their free throw percentage as the season progresses
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