Why Do You Think Mr. Baird Believes That Ap Stats?
Ap Stats:
76% of Naperville students have been alcohol free in the last 30 days. Mr. Baird believes that AP stats students are different.
Write the hypothesis statements for his belief. Assume a= .1
Some of his students take a random sample of students and comes with a 90% CI of (.80, .90)
Based on these results, what will the results of Mr. Baird’s test be? Explain.
H0: p=.76
Ha=p=/=.76
Your answer should include:
Reference that a 90% Confidence interval corresponds to a 2 sided test with a =.1
Indicating that the interval does not include .76
Diagram tends to help with explanationaaaaaaaaaaaaaaaaaaaaaaa
Test statistic = (statistic – parameter)/(standard deviation of statistic) t = (xbar-unot)/(sx/sqrt(n)) from example
equaled 1.54 go to df=14 row t statistic falls between values 1.345 and 1.761, the uppertail probability p is between .10 and .05 pvalue for this test is between .05 and .1 because the pvalue exceeds our default a = .05 significance level, we can’t conclude that the company’s new AAA batteries last longer than 30 hours, on average
using table b wisely table b gives a range of possible p values for a significance, we can still draw a conclusion from the test in much the same way as if we had a single probablility by comparing the range of possible pvalues to our deired significance level table b has other limitation for finding pvalues. It includes prob only for t distributions with df from 1 to 30 and then skips df=
H0: u = 10
Ha: u>10
N=75
t=2.333 -70…but no 70 so look at 60
Pvalue= between .01 and .02
Ho: u=10
Ha: u=/=10
N=10
t=-.51 pvalue= at least .5
calculator: 2nd dist tcdf (left, right, df)
.0112 .6223
76% of Naperville students have been alcohol free in the last 30 days. Mr. Baird believes that AP stats students are different.
Write the hypothesis statements for his belief. Assume a= .1
Some of his students take a random sample of students and comes with a 90% CI of (.80, .90)
Based on these results, what will the results of Mr. Baird’s test be? Explain.
H0: p=.76
Ha=p=/=.76
Your answer should include:
Reference that a 90% Confidence interval corresponds to a 2 sided test with a =.1
Indicating that the interval does not include .76
Diagram tends to help with explanationaaaaaaaaaaaaaaaaaaaaaaa
Test statistic = (statistic – parameter)/(standard deviation of statistic) t = (xbar-unot)/(sx/sqrt(n)) from example
equaled 1.54 go to df=14 row t statistic falls between values 1.345 and 1.761, the uppertail probability p is between .10 and .05 pvalue for this test is between .05 and .1 because the pvalue exceeds our default a = .05 significance level, we can’t conclude that the company’s new AAA batteries last longer than 30 hours, on average
using table b wisely table b gives a range of possible p values for a significance, we can still draw a conclusion from the test in much the same way as if we had a single probablility by comparing the range of possible pvalues to our deired significance level table b has other limitation for finding pvalues. It includes prob only for t distributions with df from 1 to 30 and then skips df=
H0: u = 10
Ha: u>10
N=75
t=2.333 -70…but no 70 so look at 60
Pvalue= between .01 and .02
Ho: u=10
Ha: u=/=10
N=10
t=-.51 pvalue= at least .5
calculator: 2nd dist tcdf (left, right, df)
.0112 .6223