In: Statistics and Probability
The big idea behind hypothesis testing is that we have an assumption about reality, and we see if the data fits that assumption. The whole process gets complicated by all the notation and calculations, but essentially we’re deciding if the assumption is possible, or if the data leads us to reject it.
1. Your friend Hamad claims to be exceptional at basketball and can make 90% of free throws. You watch him at the gym for a week and find out that he makes 64 shots out of 176 attempts.
Don’t do any calculations here. Just give a quick look at his results and claim, and then a statement about whether it backs up Hamad’s claim. Please write a complete sentence (or more).
2. In each situation, determine whether you should reject or fail to reject H0.
a. p-value = 0.15, α = 0.10 b. p-value = 0.015, α = 0.02
c. z = 2.345, critical value = 1.645 d. test statistic = -2.56, critical value = 1.96
3. Yolanda thinks she can roll a 1 on a 6-sided die more often than chance would predict. Write hypotheses to test this. Be sure to define p in words (p = the proportion of…).
4. Test 2: The Japanese harvester beetle has infected several forests in the Northwest. Official estimates are that 17% of trees are infected. You are a park ranger who has been seeing a lot of these beetles lately, and you think the rate is higher in your area. You check 400 trees around your cabin and find that 79 of them are infected.
1.
It doesn't back up Hamad's claim because he claims to make 90% of free throws whereas the result showed that he did not even make 50% of free throws out of 176 attempts.
2.
a.
p-value: 0.15 > α: 0.10; Failed to reject H0.
b.
p-value: 0.015 < α: 0.02; Reject H0.
c.
z: 2.345 > critical value: 1.645; Reject H0.
d.
Absolute value of test statistic: 2.56 > critical value = 1.96; Reject H0.
3.
Null Hypothesis(H0):
The proportion of getting 1 is not significantly greater than 1/6.
p 1/6
AlternativeHypothesis(H1):
The proportion of getting 1 is significantly greater than 1/6.
p > 1/6
4.
Step 1:
Null Hypothesis(H0):
The proportion of infected trees is not significantly greater than 17%.
p 0.17
Alternative Hypothesis(H1):
The proportion of infected trees is significantly greater than 17%.
p > 0.17 (Right tailed test)
Step 2:
Sample proportion, =x/n =79/400 =0.1975 =19.75%
Standard Error, SE = = =0.02
Test statistic, Z =( - p)/SE =(0.1975 - 0.17)/0.02 =1.375
Step 3:
The critical value of Z at α = 0.03 for a right tailed test is Zcrit =1.88
Step 4:
Conclusion:
Since Z: 1.375 < Zcrit :1.88, we failed to reject the null hypothesis(H0) at 0.03 significance level.
Thus, we do not have enough evidence to claim that the proportion of infected trees is significantly greater than 17%.
Conditions:
Since the sample size, n is large enough (n =400 > 30), we can use Z-score.