1. Chapter 9: testing a claim
Ch. 9 Introduction

2. 𝑝→ true proportion of made free throws by Mr. B
𝑝 =
=0.66
𝑝=0.80
𝑝<0.80
Assuming Mr. Brinkhus’ claim is true, how likely is it that he would
get a 𝑝 of 0.66 or lower purely by chance?

3. Each of you will simulate 50 free throws using the RandInt function on your calculator.
Do 50 repetitions (50 free throws)
Use RandInt(0, 9, 50)
0 – 7  made shot
8 – 9  missed shot
When you get your 𝑝 value, plot it on the class dot plot.
How likely is it that Mr. Brinkhus makes 66% or less of 50 free throws?

4. Now let’s use a TRUE Sampling Distribution of 𝑝 .
Normal?
YES!
Independent?
YES!
Check: 𝑛𝑝≥10
𝑛 1−𝑝 ≥10
Check: 10% condition
→40≥10
→10≥10
10 50 =500
< thousands
𝜎 𝑝 = 𝑝(1−𝑝) 𝑛
= =.057
Sampling Distribution of 𝑝
The probability of Mr. Brinkhus shooting 66% or lower of his free throws given he claims to shoot 80% is 0.7%, which is highly unlikely.
This gives convincing evidence that his claim is not true.
N(0.8, .057)
0.66
0.8
normalcdf −99999, 0.66, 0.8, .057 =.007
𝑝-value
P( 𝑝 ≤ 𝑝=.8

5. true mean, 𝜇, of the age of pennies in the container
𝑥 =
19.5
𝜇=22
Null Hypothesis 𝐻 0 :𝜇=22
𝜇≠22
Alternative Hypothesis 𝐻 𝑎 :𝜇≠22
This time we have a two-sided alternative hypothesis.
Before was a one-sided alternative hypothesis 𝑝<.80 .
Assuming Mr. Brinkhus’ claim is true, how likely is it that he would
get an 𝑥 of 19.5 or lower (or 24.5 or higher) purely by chance?
Do we know 𝜎?
No!

6. Remember, since we don’t know 𝜎, we’re using a 𝒕-distribution!
Normal?
Independent?
YES!
Since 𝑛=40≥30, the sampling distribution of 𝑥 is approximately normal.
Check: 10% condition
10 40 =400
< a lot
= =.474
𝜎 𝑥 = 𝑠 𝑥 𝑛
Sampling Distribution of 𝑥
df =𝑛−1=40−1=39
N(22, .474)
𝑡= 19.5− =−5.27
area =0⋅2=0
19.5 22
24.5
𝑝-value
𝑡𝑐𝑑𝑓(𝑙𝑜𝑤𝑒𝑟, 𝑢𝑝𝑝𝑒𝑟, 𝑑𝑓)
=𝑡cdf −9999,−5.27, 39 =0
Very unlikely to occur purely by chance, so we have strong evidence against Mr. Brinkhus’ claim.