This is a question I got partially incorrect on my last exam, I posted this already and still have gotten no responds. I really need help with this!

Question

In basketball, a player gets two foul shots when fouled. In a random sample of games over Larry Bird’s long NBA career, he hit the second foul shot in 45 of 53 attempts after the first was missed and hit the second foul shot 220 of 279 times after the first shot was made. Is there sufficient evidence to say that the probability that Bird made the second free throw is different on whether he made the first free throw or missed it? Also construct a 95% confidence interval for the difference between his free throw shooting. 

Test: which test are you using? Identify it by name

 

State the Given information (p,q,n for both proportions):

 

Parameter of interest? State in words what we are looking for here. 

 

Conditions: List and state the conditions here? Are they met? Why or why not? Show work for np and nq for both proportions.

 

State the Null and Alternative Hypothesis using symbols and words

 

Calculate the critical value also known as the test statistic. You may use your calculator to run the test at this point to get this value.

 

P-value:________? 

 

Conclusion? Reject or fail to reject the null? 

 

Conclusion in words? What does this mean in context of the question?

 

Confidence interval: show work and state your final interval using interval notation. (____________ , _____________) 

 

Interpret your confidence interval in words. Does it support your conclusion from 8 and 9? Why or why not? 

Answer 1

State: We want to test the following hypotheses

H₀: p₁ - p₂ = 0 Null hypothesis

Ha: p1 - p2≠0 Alternative hypothesis

Where p₁ = the proportion of Larry Bird's 2nd free throws shots made when he made his first, and p₂ = the proportion of 2nd free throws shots Bird made after he made his first.

Here, p_{1 hat} = 45/53 and p_{2 hat} = 220/279. We will use a pooled proportion, q, which is equal to the following:

q = (45+220)/(53+279) = 0.798

I am assuming the notation of q is the pooled proportion in this case. The pooled sample proportion can be used here as we are pulling both proportions from one large sample. We also use the 'hats' here as we are speaking of specific samples.

The difference in proportions given with H₀ and H_a will provide the information necessary to make a correct conclusion.

Plan: Using a significance level of α=0.05, we will perform a 2-proportion-z-test given the following conditions are met: I use this significance level as it will relate directly with a 95% Con. Interval later.

Random: The data were produced using two independent random samples.

"Random sample of games listed in question"

10% Condition: We need to check if n₁≤(1/10)N₁ and n₂≤(1/10)N₂

True

Large Counts: np≥ 10 and n(1-p)≥10, where n is the number of sample sizes.

n₁q≥10 n₁(1-q)≥10 n₂q≥10 n₂(1-q)≥10

(53)(0.798)≥10 (53)(1-0.798)≥10 (279)(0.798)≥10 (279)(1-0.798)≥10

42.3≥10 10.7≥10 222.6≥10 56.4≥10

Met Met Met Met

Do: Using the calculator, we can use the 2PropZTest. To get here, you go to Stat>Tests>2PropZTest. Depending on your calculator, your options will be different. These are the options for a TI-84 CE calculator.

x1: 45 This is the number of 2nd free throws made after missing the first.

n1: 53

x2: 220 This is the number of 2nd free throws made after making the first.

n2: 279

p₁≠p₂ We use this option as we move p₂ over to the other side of the = in the hypotheses.

This should give the following results:

z = 1.00645 **This is your critical value**

p = 0.3142 **This is our p-value**

p_{1 hat} = 0.849 **This is the original proportion for p_{1 hat}, only in decimal form**

p_{2 hat} = 0.78853 **This is the original proportion for p_{2 hat}, only in decimal form**

p_hat = 0.79819 ** This is our q value, or the pooled proportion, which was calculated earlier**

n1: 53

n2: 279

Conclusion: Since our p-value (0.3142) is greater than our significance level (0.05), we fail to reject the null hypothesis (H₀). Meaning, we do not have convincing evidence that the probability that Bird made the second free throw is different on whether he made the first free throw or missed it.

Confidence Interval: As all of the criteria for a 95% confidence has already been proven to be met, we can go straight to calculations:

Confidence Interval = (p_{1 hat} - p_{2 hat})±z^∗(Standard Error) where z^∗ is the critical value, (1.96 in this case) and the standard error is as follows:

SE = √[(p_{1 hat}(1-p_{1 hat})/n₁)+(p_{2 hat}(1-p_{2 hat})/n₂)], which in this case is 0.0549.

When we plug everything into the Conf. Interval equation, we see that

[(45/53) - (220/279)] ± (1.96)(0.0549)

0.06053 ± 0.107604

(-0.047074, 0.27143)

Since zero is included in this interval, we reject the null hypothesis, H₀. We arrive at the same conclusion from the 2PropZTest: We do not have convincing evidence that the probability that Bird made the second free throw is different on whether he made the first free throw or missed it.

**Note: You could also do the confidence interval in calculator using the 2PropZInt function. You get there by going to Stat> Tests> 2PropZInt.

I hope this helps! Let me know if you need any further clarification.

Harrison